Chapter 1 Number System

The given expression is a sum of place values multiplied by digits.

Let's evaluate each term in the expression:

$3 \times 10000 = 30000$

$0 \times 1000 = 0$

$8 \times 100 = 800$

$0 \times 10 = 0$

$7 \times 1 = 7$

Now, we sum these values:

$30000 + 0 + 800 + 0 + 7 = 30807$

The value of the expression $3 \times 10000 + 0 \times 1000 + 8 \times 100 + 0 \times 10 + 7 \times 1$ is 30807.

Comparing this value with the given options, we find that it matches option (B).

The correct option is (B).

We need to determine which of the given options is equivalent to 1 billion.

Let's define the values of a billion, a million, and a lakh according to the international and Indian number systems:

1 billion = $1,000,000,000$ (One thousand million) in the international system.

1 million = $1,000,000$ (Ten lakh) in the international and Indian systems respectively.

1 lakh = $100,000$ in the Indian system.

Now let's evaluate each option:

(A) 100 millions

$100 \times 1,000,000 = 100,000,000$

This is equal to 100 million, which is $0.1$ billion.

(B) 10 millions

$10 \times 1,000,000 = 10,000,000$

This is equal to 10 million, which is $0.01$ billion.

(C) 1000 lakhs

$1000 \times 100,000 = 100,000,000$

This is equal to 100 million, which is $0.1$ billion.

(D) 10000 lakhs

$10000 \times 100,000 = 1,000,000,000$

This is equal to 1 billion.

Comparing the results, we find that 1 billion is equal to 10000 lakhs.

The correct option is (D).

We need to identify the Roman numeral that is incorrectly formed based on the rules of Roman numerals.

Let's examine each option:

(A) LXII:

L represents 50.

X represents 10.

I represents 1.

II represents 1 + 1 = 2.

LXII represents 50 + 10 + 2 = 62.

This is a correctly formed Roman numeral.

(B) XCI:

X placed before C represents subtraction: $100 - 10 = 90$.

I represents 1.

XCI represents 90 + 1 = 91.

This is a correctly formed Roman numeral (X can be subtracted from L and C).

(C) LC:

L represents 50.

C represents 100.

If this were addition, it would be 50 + 100 = 150, which is represented as CL.

If this were subtraction, L (50) is placed before C (100). However, the symbol L (50) is never used for subtraction in Roman numerals. Only I, X, and C can be placed before symbols of larger value for subtraction (I before V and X; X before L and C; C before D and M).

Therefore, LC is an incorrectly formed Roman numeral.

(D) XLIV:

X placed before L represents subtraction: $50 - 10 = 40$.

I placed before V represents subtraction: $5 - 1 = 4$.

XLIV represents 40 + 4 = 44.

This is a correctly formed Roman numeral.

Based on the analysis, the incorrect Roman numeral is LC.

The correct option is (C).

The question asks which of the given mathematical expressions is not defined.

Let's examine each option:

(A) $5 + 0$

This is an addition operation. Adding 0 to any number results in the number itself.

$5 + 0 = 5$

This operation is well-defined.

(B) $5 - 0$

This is a subtraction operation. Subtracting 0 from any number results in the number itself.

$5 - 0 = 5$

This operation is well-defined.

(C) $5 \times 0$

This is a multiplication operation. Multiplying any number by 0 results in 0.

$5 \times 0 = 0$

This operation is well-defined.

(D) $5 \div 0$

This is a division operation where the divisor is 0. Division by zero is undefined in mathematics. This is because if we assume $\frac{5}{0} = k$ for some number $k$, then by the definition of division, we would have $0 \times k = 5$. However, $0 \times k = 0$ for any finite number $k$, so there is no number $k$ that satisfies $0 \times k = 5$.

Therefore, division by zero is not defined.

Based on the evaluation of each option, the expression that is not defined is $5 \div 0$.

The correct option is (D).

Let the non-zero whole number be $n$. Since it is a non-zero whole number, $n$ belongs to the set $\{1, 2, 3, ...\}$.

The successor of $n$ is $n+1$.

The product of the non-zero whole number and its successor is $n \times (n+1)$, which can be written as $n(n+1)$.

We need to determine which of the given numbers ($2, 3, 4, 5$) always divides the product $n(n+1)$.

Consider two consecutive integers, $n$ and $n+1$. One of these integers must be an even number, i.e., divisible by 2.

Case 1: If $n$ is even, then $n$ is divisible by 2. Since $n(n+1)$ has $n$ as a factor, the product $n(n+1)$ is divisible by 2.

Case 2: If $n$ is odd, then $n+1$ must be even. Since $n(n+1)$ has $n+1$ as a factor, and $n+1$ is divisible by 2, the product $n(n+1)$ is divisible by 2.

In both cases, the product of a non-zero whole number and its successor is always divisible by 2.

Let's check if it is always divisible by other options:

If $n=1$, the product is $1 \times 2 = 2$. This is divisible by 2, but not by 3, 4, or 5.

If $n=2$, the product is $2 \times 3 = 6$. This is divisible by 2 and 3, but not by 4 or 5.

If $n=3$, the product is $3 \times 4 = 12$. This is divisible by 2, 3, and 4, but not by 5.

If $n=4$, the product is $4 \times 5 = 20$. This is divisible by 2, 4, and 5, but not by 3.

From these examples, we can see that the product is not always divisible by 3, 4, or 5.

However, the product $n(n+1)$ is always divisible by 2.

The correct option is (A).

We need to find the number of factors of 36.

A factor of 36 is a number that divides 36 completely without leaving any remainder.

We can list the factors of 36 by checking which whole numbers divide 36 evenly, starting from 1:

• $36 \div 1 = 36$. So, 1 and 36 are factors.
• $36 \div 2 = 18$. So, 2 and 18 are factors.
• $36 \div 3 = 12$. So, 3 and 12 are factors.
• $36 \div 4 = 9$. So, 4 and 9 are factors.
• $36 \div 5$ is not a whole number.
• $36 \div 6 = 6$. So, 6 is a factor (we only list it once).

Once we reach a factor where the quotient is less than or equal to the divisor (like 6 and 6), we have found all pairs of factors.

The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

Let's count the number of factors:

There are 9 factors.

Alternatively, using prime factorization:

First, find the prime factorization of 36.

$\begin{array}{c|cc} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $36 = 2^2 \times 3^2$.

To find the number of factors, we add 1 to each exponent in the prime factorization and multiply the results.

Number of factors $= (2+1) \times (2+1) = 3 \times 3 = 9$.

Both methods show that the number of factors of 36 is 9.

The correct option is (D).

We need to find the sum of the first three common multiples of 3, 4, and 9.

First, we find the Least Common Multiple (LCM) of 3, 4, and 9. The common multiples of these numbers are the multiples of their LCM.

Let's find the prime factorization of each number:

Prime factorization of 3 is 3.

Prime factorization of 4 is $2 \times 2 = 2^2$.

Prime factorization of 9 is $3 \times 3 = 3^2$.

To find the LCM, we take the highest power of all prime factors involved in the factorizations:

LCM$(3, 4, 9) = 2^2 \times 3^2 = 4 \times 9 = 36$.

The common multiples of 3, 4, and 9 are the multiples of 36.

The multiples of 36 are $36 \times 1$, $36 \times 2$, $36 \times 3$, $36 \times 4$, ...

These are 36, 72, 108, 144, ...

The first three common multiples of 3, 4, and 9 are 36, 72, and 108.

Now, we find the sum of these three common multiples:

Sum $= 36 + 72 + 108$

Sum $= 108 + 108$

Sum $= 216$

The sum of the first three common multiples of 3, 4, and 9 is 216.

The correct option is (D).

In the Indian System of Numeration, the place value chart is as follows:

Ones, Tens, Hundreds, Thousands, Ten Thousands, Lakhs, Ten Lakhs, Crores, Ten Crores, ...

Commas are placed as follows:

• The first comma comes after the Hundreds place (3 digits from the right).
• Subsequent commas come after every two digits, marking the Lakhs and Crores periods.

Let's apply this to the number 61711682:

Starting from the rightmost digit (2):

• The first comma is placed after 3 digits: 61711,682
• The second comma is placed after the next 2 digits: 617,11,682
• The third comma is placed after the next 2 digits: 6,17,11,682

So, the number 61711682 is written as 6,17,11,682 in the Indian System of Numeration.

The complete statement is: In Indian System of Numeration, the number 61711682 is written, using commas, as 6,17,11,682.

We are looking for the smallest 4-digit number where all the digits are different.

To form the smallest possible number with a fixed number of digits, we should arrange the smallest available distinct digits in ascending order from left to right (highest place value to lowest place value).

We need four different digits. The smallest digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

For a 4-digit number, the first digit (at the thousands place) cannot be 0.

So, the smallest possible digit for the thousands place is 1.

The remaining three digits must be different from 1 and also different from each other. To keep the number smallest, we should use the smallest available digits for the subsequent places.

The smallest digit available other than 1 is 0. So, the digit for the hundreds place is 0.

The smallest digit available other than 1 and 0 is 2. So, the digit for the tens place is 2.

The smallest digit available other than 1, 0, and 2 is 3. So, the digit for the ones place is 3.

Arranging these digits in order: Thousands, Hundreds, Tens, Ones.

Digits are 1, 0, 2, 3.

The number formed is 1023.

This is a 4-digit number (since the first digit is not 0), and all its digits (1, 0, 2, 3) are different.

Any other combination of four different digits would result in a larger number (e.g., starting with 2, or using larger digits in earlier places).

The smallest 4 digit number with different digits is 1023.

The complete statement is: The smallest 4 digit number with different digits is 1023.

A factor of a number is any number that divides it evenly, leaving no remainder.

Let's consider the number of factors for different types of whole numbers:

• Numbers with exactly two factors (1 and the number itself) are called prime numbers (e.g., 2, 3, 5, 7).
• The number 1 has only one factor (1) and is neither prime nor composite.
• Numbers with more than two factors are called composite numbers (e.g., 4 (factors 1, 2, 4), 6 (factors 1, 2, 3, 6), 8 (factors 1, 2, 4, 8), 9 (factors 1, 3, 9)).

The question asks for the name given to numbers having more than two factors.

Based on the definitions, numbers with more than two factors are called composite numbers.

The complete statement is: Numbers having more than two factors are called composite numbers.

To round a number to the nearest hundred, we look at the digit in the tens place.

In the number $58963$, the digit in the tens place is $6$.

Since the tens digit ($6$) is $5$ or greater, we round up the hundreds digit.

The hundreds digit is $9$. Rounding $9$ up requires carrying over $1$ to the thousands place.

$58963$ rounded to the nearest hundred becomes $59000$.

The given statement says the number $58963$ rounded off to nearest hundred is $58900$.

This statement is False.

We need to convert the Roman numerals to their Hindu-Arabic equivalents.

The Roman numeral $L$ represents $50$.

The Roman numeral $X$ represents $10$.

The Roman numeral $V$ represents $5$.

The Roman numeral $I$ represents $1$.

For $LXXV$:

$LXXV = L + X + X + V = 50 + 10 + 10 + 5 = 75$.

For $LXXIV$:

In $IV$, $I$ is placed before $V$, which means $I$ is subtracted from $V$.

$IV = V - I = 5 - 1 = 4$.

So, $LXXIV = L + X + X + IV = 50 + 10 + 10 + 4 = 74$.

Now we compare the values: $75$ and $74$.

Clearly, $75$ is greater than $74$.

Therefore, $LXXV$ is greater than $LXXIV$.

The given statement is True.

The statement consists of two parts connected by "So,". Let's examine each part.

The first part says: "If a number is divisible by 2 and 3, then it is also divisible by 6."

This statement is True. A number is divisible by $6$ if and only if it is divisible by both $2$ and $3$, because $2$ and $3$ are coprime factors of $6$.

The second part says: "if a number is divisible by 2 and 4, it must be divisible by 8."

This statement is False. While $8$ is a multiple of both $2$ and $4$, being divisible by $2$ and $4$ (which is equivalent to being divisible by $4$) does not guarantee divisibility by $8$. This is because $2$ and $4$ are not coprime factors of $8$. The prime factors of $8$ are $2 \times 2 \times 2$ ($2^3$). Divisibility by $4$ only ensures divisibility by $2^2$.

Consider the number $12$.

$12$ is divisible by $2$ ($12 \div 2 = 6$).

$12$ is divisible by $4$ ($12 \div 4 = 3$).

However, $12$ is not divisible by $8$ ($12 \div 8 = 1$ with a remainder of $4$).

Since the second part of the overall statement is false, the entire statement is False.

Given:

Population of Agra district in the year 2001 = $36,20,436$.

Population of Aligarh district in the year 2001 = $29,92,286$.

To Find:

Total population of the two districts in that year.

Solution:

The total population of the two districts is the sum of the individual populations.

Total Population = Population of Agra + Population of Aligarh

Total Population = $36,20,436 + 29,92,286$

Let's perform the addition:

$\begin{array}{ccccccc} & 3 & 6 & 2 & 0 & 4 & 3 & 6 \\ + & 2 & 9 & 9 & 2 & 2 & 8 & 6 \\ \hline & 6 & 6 & 1 & 2 & 7 & 2 & 2 \\ \hline \end{array}$

So, the total population is $66,12,722$.

The total population of Agra and Aligarh districts in the year 2001 was $66,12,722$.

(i) Rounding off to the nearest tens

To round $5981$ to the nearest ten, we look at the units digit, which is $1$. Since $1 < 5$, we round down.

$5981$ rounded to the nearest ten is $5980$.

To round $4428$ to the nearest ten, we look at the units digit, which is $8$. Since $8 \geq 5$, we round up.

$4428$ rounded to the nearest ten is $4430$.

Estimated product = $5980 \times 4430$.

Estimated product = $26491400$.

(ii) Rounding off to the nearest hundreds

To round $5981$ to the nearest hundred, we look at the tens digit, which is $8$. Since $8 \geq 5$, we round up.

$5981$ rounded to the nearest hundred is $6000$.

To round $4428$ to the nearest hundred, we look at the tens digit, which is $2$. Since $2 < 5$, we round down.

$4428$ rounded to the nearest hundred is $4400$.

Estimated product = $6000 \times 4400$.

Estimated product = $26400000$.

We need to find the product of $8739$ and $102$ using the distributive property.

The distributive property states that $a \times (b + c) = a \times b + a \times c$.

We can rewrite the number $102$ as a sum of two numbers, such as $100 + 2$.

So, we can write the product as:

$8739 \times 102 = 8739 \times (100 + 2)$

Applying the distributive property:

$8739 \times (100 + 2) = (8739 \times 100) + (8739 \times 2)$

Now, we calculate each part:

$8739 \times 100 = 873900$

$8739 \times 2$:

$\begin{array}{cc}& 8 & 7 & 3 & 9 \\ \times & & & & 2 \\ \hline 1 & 7 & 4 & 7 & 8 \\ \hline \end{array}$

So, $8739 \times 2 = 17478$.

Now, we add the two results:

$873900 + 17478$

$\begin{array}{ccccccc} & 8 & 7 & 3 & 9 & 0 & 0 \\ + & & 1 & 7 & 4 & 7 & 8 \\ \hline & 8 & 9 & 1 & 3 & 7 & 8 \\ \hline \end{array}$

Therefore, the product $8739 \times 102$ using the distributive property is $891378$.

Given:

Dimensions of the floor = $4.5$ metres $\times 3$ metres.

The floor is rectangular with length $4.5$ m and width $3$ m.

To Find:

Minimum number of complete square marble slabs of equal size required to cover the entire floor.

Solution:

To cover the floor with complete square marble slabs of equal size, the side length of the square slab must be a common factor of the length and width of the floor.

To find the minimum number of slabs, the side length of the square slab must be the greatest common factor (HCF) of the length and width of the floor.

First, let's convert the dimensions from metres to centimetres to work with whole numbers.

$1$ metre = $100$ centimetres.

Length of the floor = $4.5$ m = $4.5 \times 100$ cm = $450$ cm.

Width of the floor = $3$ m = $3 \times 100$ cm = $300$ cm.

Now, we find the HCF of $450$ and $300$. We can use the prime factorization method.

Prime factorization of $450$:

$\begin{array}{c|cc} 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $450 = 2 \times 3 \times 3 \times 5 \times 5 = 2^1 \times 3^2 \times 5^2$.

Prime factorization of $300$:

$\begin{array}{c|cc} 2 & 300 \\ \hline 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $300 = 2 \times 2 \times 3 \times 5 \times 5 = 2^2 \times 3^1 \times 5^2$.

The HCF is found by taking the lowest power of each common prime factor.

Common prime factors are $2, 3, 5$.

Lowest power of $2$ is $2^1$.

Lowest power of $3$ is $3^1$.

Lowest power of $5$ is $5^2$.

HCF$(450, 300) = 2^1 \times 3^1 \times 5^2 = 2 \times 3 \times 25 = 6 \times 25 = 150$.

The side length of the largest possible square marble slab is $150$ cm.

Now, we calculate the number of slabs needed along the length and width.

Number of slabs along the length = $\frac{\text{Length of floor}}{\text{Side length of slab}} = \frac{450 \text{ cm}}{150 \text{ cm}} = 3$.

Number of slabs along the width = $\frac{\text{Width of floor}}{\text{Side length of slab}} = \frac{300 \text{ cm}}{150 \text{ cm}} = 2$.

Total number of slabs required = (Number along length) $\times$ (Number along width) = $3 \times 2 = 6$.

Thus, the minimum number of complete square marble slabs of equal size required to cover the entire floor is $6$.

The given number is $428721$.

The digit $2$ appears at two places in the number.

The first $2$ is in the ten thousands place.

Its place value is $2 \times 10000 = 20000$.

The second $2$ is in the tens place.

Its place value is $2 \times 10 = 20$.

The product of the place values of the two 2’s is the product of $20000$ and $20$.

Product = $20000 \times 20 = 400000$.

Comparing this value with the given options:

(A) $4$

(B) $40000$

(C) $400000$

(D) $40000000$

The product matches option (C).

The correct answer is (C) 400000.

We need to evaluate the given expression: $3 \times 10000 + 7 \times 1000 + 9 \times 100 + 0 \times 10 + 4$.

Let's calculate each term:

$3 \times 10000 = 30000$

$7 \times 1000 = 7000$

$9 \times 100 = 900$

$0 \times 10 = 0$

$4 = 4$

Now, we add these values together:

$30000 + 7000 + 900 + 0 + 4$

$\begin{array}{cccccc} & 3 & 0 & 0 & 0 & 0 \\ & & 7 & 0 & 0 & 0 \\ & & & 9 & 0 & 0 \\ & & & & 0 & 0 \\ + & & & & & 4 \\ \hline & 3 & 7 & 9 & 0 & 4 \\ \hline \end{array}$

The sum is $37904$.

Comparing the result with the given options:

(A) $3794$

(B) $37940$

(C) $37904$

(D) $379409$

The result $37904$ matches option (C).

The correct answer is (C) 37904.

The greatest 7-digit number is the number with $7$ nines.

Greatest 7-digit number = $9,999,999$.

We are asked to find the number obtained by adding $1$ to the greatest 7-digit number.

$9,999,999 + 1$.

Let's perform the addition:

$\begin{array}{ccccccc} & 9 & 9 & 9 & 9 & 9 & 9 & 9 \\ + & & & & & & & 1 \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}$

The result is $10,000,000$.

Now, let's look at the options and their values:

(A) 10 thousand = $10,000$

(B) 1 lakh = $100,000$

(C) 10 lakh = $1,000,000$ (which is also 1 million)

(D) 1 crore = $10,000,000$ (which is also 10 million)

The result $10,000,000$ is equal to 1 crore.

The correct answer is (D) 1 crore.

The given number is $9578$.

To write the expanded form, we represent the number as the sum of the place values of its digits.

The digit $9$ is in the thousands place, so its place value is $9 \times 1000$.

The digit $5$ is in the hundreds place, so its place value is $5 \times 100$.

The digit $7$ is in the tens place, so its place value is $7 \times 10$.

The digit $8$ is in the units place, so its place value is $8 \times 1$.

So, the expanded form of $9578$ is $9 \times 1000 + 5 \times 100 + 7 \times 10 + 8 \times 1$.

Comparing this with the given options:

(A) $9 \times 10000 + 5 \times 1000 + 7 \times 10 + 8 \times 1 = 90000 + 5000 + 70 + 8 = 95078$ (Incorrect)

(B) $9 \times 1000 + 5 \times 100 + 7 \times 10 + 8 \times 1 = 9000 + 500 + 70 + 8 = 9578$ (Correct)

(C) $9 \times 1000 + 57 \times 10 + 8 \times 1 = 9000 + 570 + 8 = 9578$ (Correct form but not the standard expanded form by individual digits)

(D) $9 \times 100 + 5 \times 100 + 7 \times 10 + 8 \times 1 = 900 + 500 + 70 + 8 = 1478$ (Incorrect)

The standard expanded form expressing each digit's place value is given in option (B).

The correct answer is (B) 9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1.

To round a number to the nearest thousands, we look at the digit in the hundreds place.

The given number is $85642$.

The digit in the hundreds place is $6$.

Since the hundreds digit ($6$) is $5$ or greater, we round up the thousands digit.

The thousands digit is $5$. Rounding up $5$ makes it $6$.

All digits to the right of the thousands place become $0$.

So, $85642$ rounded off to the nearest thousands is $86000$.

Comparing this value with the given options:

(A) $85600$

(B) $85700$

(C) $85000$

(D) $86000$

The rounded number matches option (D).

The correct answer is (D) 86000.

We are given the digits $5, 9, 2, 6$.

We need to form the largest 4-digit number using these digits, such that exactly one digit is used twice.

The available digits are $2, 5, 6, 9$. To form the largest number, we should use the largest possible digits in the higher place value positions (thousands, hundreds).

Since we need to repeat one digit, and we want the largest possible number, we should repeat the largest available digit, which is $9$.

So, the digits we will use are $9, 9$ (using 9 twice) and the remaining two largest unique digits from the given set, which are $6$ and $5$.

The four digits to form the number are $9, 9, 6, 5$.

To make the largest number from these digits, we arrange them in descending order: $9965$.

Let's check if repeating other digits would yield a larger number:

- Repeating $6$: Digits are $9, 6, 6, 5$. Largest number is $9665$.

- Repeating $5$: Digits are $9, 6, 5, 5$. Largest number is $9655$.

- Repeating $2$: Digits are $9, 6, 5, 2$ (and $2$ used twice, but we need 4 digits in total, so $9, 6, 5, 2$ plus one more repetition. If we repeat $2$, the digits could be $9, 6, 5, 2$ using the original set once, and then picking one more from the set $2, 5, 6, 9$ to repeat. To make it largest, we repeat the largest digit 9. So the digits are $9,9,6,5$ as determined above.).

Among $9965, 9665, 9655$, the largest is $9965$.

Comparing this value with the given options:

(A) $9652$ (Digits are $9, 6, 5, 2$. No repetition)

(B) $9562$ (Digits are $9, 5, 6, 2$. No repetition)

(C) $9659$ (Digits are $9, 6, 5, 9$. Digit $9$ is repeated)

(D) $9965$ (Digits are $9, 9, 6, 5$. Digit $9$ is repeated)

Both (C) and (D) use one digit twice. The number in (D), $9965$, is larger than the number in (C), $9659$.

The correct answer is (D) 9965.

The given number is $58695376$.

In the Indian System of Numeration, the periods are grouped from right to left as follows:

Units (first 3 digits), Thousands (next 2 digits), Lakhs (next 2 digits), Crores (next 2 digits), and so on.

We write the number $58695376$ and place commas according to the Indian system grouping.

Starting from the right:

The first period (Units) consists of the last three digits: $376$.

The second period (Thousands) consists of the next two digits: $95$.

The third period (Lakhs) consists of the next two digits: $86$.

The fourth period (Crores) consists of the remaining digits: $5$.

Placing the commas after each period (from right to left, skipping the Units period's end):

$5, 86, 95, 376$.

This is read as Five Crore Eighty-six Lakh Ninety-five Thousand Three Hundred Seventy-six.

Comparing this format with the given options:

(A) $58, 69, 53, 76$ (Incorrect grouping)

(B) $58, 695, 376$ (Incorrect grouping; this format is similar to the International System for the last two groups)

(C) $5, 86, 95, 376$ (Correct grouping for the Indian System)

(D) $586, 95, 376$ (Incorrect grouping)

The representation $5, 86, 95, 376$ matches option (C).

The correct answer is (C) 5, 86, 95, 376.

We need to find the equivalent of one million in the Indian System of Numeration.

In the International System of Numeration,

One million = $1,000,000$.

In the Indian System of Numeration, the place values are grouped as Units (Hundreds, Tens, Ones), Thousands (Ten Thousands, Thousands), Lakhs (Ten Lakhs, Lakhs), Crores (Ten Crores, Crores), and so on.

Let's write $1,000,000$ in the Indian System format by placing commas according to the periods:

Starting from the right: $000$ (Units), $00$ (Thousands), $00$ (Lakhs), $1$ (Crores is not applicable here as it's less than 1 crore).

The number is $10,00,000$.

This number is read as Ten Lakh.

So, one million is equal to 10 lakh.

Comparing this with the given options:

(A) 1 lakh = $1,00,000$

(B) 10 lakh = $10,00,000$

(C) 1 crore = $1,00,00,000$

(D) 10 crore = $10,00,00,000$

Our result, $10,00,000$, matches option (B).

The correct answer is (B) 10 lakh.

To round a number to the nearest thousand, we look at the digit in the hundreds place.

- If the hundreds digit is $5$ or greater, we increase the thousands digit by $1$ and change all digits to its right (hundreds, tens, and units) to $0$.

- If the hundreds digit is less than $5$, we keep the thousands digit as it is and change all digits to its right (hundreds, tens, and units) to $0$.

We are given that the number rounded off to the nearest thousands is $5000$.

This means the original number must be in the range $[4500, 5500)$. Any number from $4500$ up to (but not including) $5500$ will round to $5000$ when rounded to the nearest thousand.

We are looking for the greatest number among the given options that falls within this range and rounds to $5000$.

Let's examine each option:

(A) $5001$: The hundreds digit is $0$. Since $0 < 5$, we round down. $5001$ rounds to $5000$.

(B) $5559$: The hundreds digit is $5$. Since $5 \geq 5$, we round up. $5559$ rounds up the thousands digit $5$ to $6$ and becomes $6000$.

(C) $5999$: The hundreds digit is $9$. Since $9 \geq 5$, we round up. $5999$ rounds up the thousands digit $5$ to $6$ and becomes $6000$.

(D) $5499$: The hundreds digit is $4$. Since $4 < 5$, we round down. $5499$ rounds to $5000$.

Options (A) and (D) both round to $5000$. We need the greatest among these.

Comparing $5001$ and $5499$, the greater number is $5499$.

Also, $5499$ is the largest integer in the range $[4500, 5500)$.

The correct answer is (D) 5499.

The given number is $6350947$.

We need to keep the digit $6$ in its original place, which is the leftmost position (the millions place in the International System or the ten lakhs place in the Indian System, depending on context, but the position is fixed).

The other digits in the number are $3, 5, 0, 9, 4, 7$. There are six other digits.

To obtain the smallest possible number by rearranging the remaining digits, we must arrange them in ascending order from the next position to the right (the hundred thousands place or lakhs place) towards the units place.

The remaining digits are $3, 5, 0, 9, 4, 7$.

Let's arrange these digits in ascending order: $0, 3, 4, 5, 7, 9$.

Now, we form the number by keeping $6$ in the first position and placing the sorted remaining digits after it:

$6034579$.

Let's compare this number with the given options:

(A) $6975430$ (Here the remaining digits are $9, 7, 5, 4, 3, 0$ - arranged in descending order, resulting in the largest number)

(B) $6043579$ (Here the remaining digits are $0, 4, 3, 5, 7, 9$)

(C) $6034579$ (Here the remaining digits are $0, 3, 4, 5, 7, 9$)

(D) $6034759$ (Here the remaining digits are $0, 3, 4, 7, 5, 9$)

Comparing options (B), (C), and (D), which all start with $60$, we look at the next digit. $3$ is the smallest among $4, 3, 3$.

Now compare (C) $6034579$ and (D) $6034759$. The next digit is $4$ in both. The next digit is $5$ in (C) and $7$ in (D). Since $5 < 7$, (C) is smaller than (D).

The smallest number formed is $6034579$.

The correct answer is (C) 6034579.

We need to check the validity of each given Roman numeral.

Recall the basic Roman numeral symbols and their values:

$I = 1$

$V = 5$

$X = 10$

$L = 50$

$C = 100$

$D = 500$

$M = 1000$

Also, recall the rules for forming Roman numerals, particularly regarding repetition and subtraction. Symbols $V, L,$ and $D$ are never repeated.

Let's evaluate each option:

(A) LXXX: This represents $L + X + X + X = 50 + 10 + 10 + 10 = 80$. This is a valid Roman numeral.

(B) LXX: This represents $L + X + X = 50 + 10 + 10 = 70$. This is a valid Roman numeral.

(C) LX: This represents $L + X = 50 + 10 = 60$. This is a valid Roman numeral.

(D) LLX: This involves the repetition of the symbol $L$. The rule states that $V, L,$ and $D$ are never repeated. Therefore, $LL$ is an invalid combination in Roman numerals.

Hence, LLX is an incorrect Roman numeral.

The correct answer is (D) LLX.

We need to form the largest 5-digit number using only three different digits.

To make the number as large as possible, we should use the largest possible digits. The largest digits are $9, 8, 7, 6, ...$.

We must use exactly three distinct digits. To maximize the number, these three digits should be the largest available digits, which are $9, 8,$ and $7$.

A 5-digit number has five places (ten thousands, thousands, hundreds, tens, units). To make the number largest, we should place the largest digit ($9$) in the leftmost positions as many times as possible.

Since we can only use three different digits ($9, 8, 7$), and we have 5 places, we must repeat at least one digit. To make the number largest, we should repeat the largest digit ($9$) the maximum number of times allowed while still using three distinct digits.

We can use $9$ three times, $8$ once, and $7$ once. The digits available are $9, 9, 9, 8, 7$.

To form the largest number with these digits, arrange them in descending order: $99987$.

This number $99987$ is a 5-digit number and uses exactly three different digits ($9, 8, 7$).

Let's examine the options and check if they meet the criteria (5 digits, exactly three different digits) and which one is the largest:

(A) $98978$: Uses digits $9, 8, 7$. Three different digits. It is a 5-digit number.

(B) $99897$: Uses digits $9, 8, 7$. Three different digits. It is a 5-digit number.

(C) $99987$: Uses digits $9, 8, 7$. Three different digits. It is a 5-digit number.

(D) $98799$: Uses digits $9, 8, 7$. Three different digits. It is a 5-digit number.

All options are 5-digit numbers using exactly three different digits ($9, 8, 7$). We need to find the largest among these.

Comparing the numbers directly:

$98978$

$99897$

$99987$

$98799$

The largest number is $99987$.

The correct answer is (C) 99987.

We need to find the smallest 4-digit number that uses exactly three different digits.

A 4-digit number starts from $1000$.

To make the number smallest, we should use the smallest possible digits and place them in the higher place value positions (thousands, hundreds, etc.) in ascending order.

The smallest digits available are $0, 1, 2, 3, ...$.

Since it's a 4-digit number, the thousands digit cannot be $0$. The smallest possible thousands digit is $1$.

We need to use exactly three different digits. To form the smallest number, these three digits should be the smallest possible digits, which are $1, 0,$ and $2$.

We have three distinct digits $(1, 0, 2)$ and need to form a 4-digit number. This means one of these three digits must be repeated. To make the number smallest, we should repeat the smallest possible digit. The smallest digit available is $0$.

So, the digits we will use are $1$ (in the thousands place), and the remaining digits will be $0, 2,$ and one more $0$ (repeating the smallest digit $0$). The set of digits is $1, 0, 0, 2$.

To form the smallest 4-digit number using these digits, we place $1$ in the thousands place and arrange the remaining digits ($0, 0, 2$) in ascending order in the hundreds, tens, and units places: $0, 0, 2$.

The number formed is $1002$.

This number $1002$ is a 4-digit number and uses exactly three different digits ($1, 0, 2$).

Let's check the given options:

(A) $1102$: Uses digits $1, 0, 2$ (three different digits).

(B) $1012$: Uses digits $1, 0, 2$ (three different digits).

(C) $1020$: Uses digits $1, 0, 2$ (three different digits).

(D) $1002$: Uses digits $1, 0, 2$ (three different digits).

All options use exactly three different digits ($1, 0, 2$). Now we compare the values to find the smallest:

$1002$

$1012$

$1020$

$1102$

The smallest number among these is $1002$.

The correct answer is (D) 1002.

We need to find the number of whole numbers that are strictly between $38$ and $68$.

Whole numbers are $0, 1, 2, 3, ...$.

The whole numbers between $38$ and $68$ are those whole numbers $n$ such that $38 < n < 68$.

These numbers are $39, 40, 41, ..., 66, 67$.

To count the number of integers between two integers $a$ and $b$ (where $a < b$), the number of integers is $b - a - 1$.

In this case, $a = 38$ and $b = 68$.

Number of whole numbers between $38$ and $68 = 68 - 38 - 1$.

$68 - 38 = 30$.

$30 - 1 = 29$.

Alternatively, the sequence of whole numbers is $39, 40, ..., 67$. This is an arithmetic progression with first term $39$, last term $67$, and common difference $1$. The number of terms is given by (Last term - First term) + 1.

Number of terms = $(67 - 39) + 1 = 28 + 1 = 29$.

There are $29$ whole numbers between $38$ and $68$.

Comparing the result with the given options:

(A) $31$

(B) $30$

(C) $29$

(D) $28$

The result $29$ matches option (C).

The correct answer is (C) 29.

The given number is $999$.

The successor of a number is the number that comes just after it.

Successor of $999 = 999 + 1 = 1000$.

The predecessor of a number is the number that comes just before it.

Predecessor of $999 = 999 - 1 = 998$.

We need to find the product of the successor and the predecessor of $999$.

Product = Successor $\times$ Predecessor

Product = $1000 \times 998$.

$1000 \times 998 = 998000$.

Comparing this result with the given options:

(A) $999000$

(B) $998000$

(C) $989000$

(D) $1998$

The calculated product $998000$ matches option (B).

The correct answer is (B) 998000.

Let the non-zero whole number be $n$.

Since it is a non-zero whole number, $n$ belongs to the set $\{1, 2, 3, 4, ...\}$.

The successor of $n$ is the number $n+1$.

We need to find the nature of the product of $n$ and its successor, which is $n \times (n+1)$.

Consider any two consecutive whole numbers, $n$ and $n+1$.

One of these two consecutive numbers must be an even number, and the other must be an odd number.

Case 1: If $n$ is even, then $n+1$ is odd. The product is Even $\times$ Odd = Even.

Case 2: If $n$ is odd, then $n+1$ is even. The product is Odd $\times$ Even = Even.

In both cases, the product of $n$ and $n+1$ is always an even number.

Let's check the options:

(A) an even number: As shown above, the product is always even.

(B) an odd number: This is false. For example, if $n=2$, the product is $2 \times 3 = 6$, which is even.

(C) a prime number: This is false. For example, if $n=2$, the product is $2 \times 3 = 6$, which is not prime. If $n=3$, the product is $3 \times 4 = 12$, which is not prime.

(D) divisible by 3: This is false. For example, if $n=4$, the product is $4 \times 5 = 20$, which is not divisible by 3.

The only property that is always true for the product of a non-zero whole number and its successor is that it is an even number.

The correct answer is (A) an even number.

Let the whole number be $x$.

When the whole number $x$ is added to $25$, the resulting number is $25 + x$.

When the same whole number $x$ is subtracted from $25$, the resulting number is $25 - x$.

We need to find the sum of these two resulting numbers.

Sum = $(25 + x) + (25 - x)$.

Using the associative and commutative properties of addition:

Sum = $25 + x + 25 - x$

Sum = $25 + 25 + x - x$

Sum = $50 + 0$

Sum = $50$.

Let's try with an example. Let the whole number be $5$.

Number added to $25 = 25 + 5 = 30$.

Number subtracted from $25 = 25 - 5 = 20$.

Sum of resulting numbers = $30 + 20 = 50$.

Let's try with another example. Let the whole number be $0$.

Number added to $25 = 25 + 0 = 25$.

Number subtracted from $25 = 25 - 0 = 25$.

Sum of resulting numbers = $25 + 25 = 50$.

The sum is always $50$, regardless of the whole number added and subtracted. Note that for the subtraction $25-x$ to result in a whole number if required, $x$ must be less than or equal to $25$. However, the question states "a whole number is added to 25 and the same number is subtracted from 25", implying the operations are performed, and we are interested in the sum of the results. The result of $25-x$ might be a negative integer if $x > 25$, but the sum $(25+x)+(25-x)$ is still $50$. Assuming "resulting numbers" are just the outcomes of the operations, the sum is always 50.

Comparing the result with the given options:

(A) $0$

(B) $25$

(C) $50$

(D) $75$

The result $50$ matches option (C).

The correct answer is (C) 50.

We need to evaluate each statement to determine which one is not true.

(A) $(7 + 8) + 9 = 7 + (8 + 9)$

This is the associative property of addition.

Left side: $(7 + 8) + 9 = 15 + 9 = 24$.

Right side: $7 + (8 + 9) = 7 + 17 = 24$.

Since $24 = 24$, this statement is True.

(B) $(7 \times 8) \times 9 = 7 \times (8 \times 9)$

This is the associative property of multiplication.

Left side: $(7 \times 8) \times 9 = 56 \times 9$.

$\begin{array}{cc}& & 5 & 6 \\ \times & & & 9 \\ \hline & 5 & 0 & 4 \\ \hline \end{array}$

So, $(7 \times 8) \times 9 = 504$.

Right side: $7 \times (8 \times 9) = 7 \times 72$.

$\begin{array}{cc}& & 7 & 2 \\ \times & & & 7 \\ \hline & 5 & 0 & 4 \\ \hline \end{array}$

So, $7 \times (8 \times 9) = 504$.

Since $504 = 504$, this statement is True.

(C) $7 + 8 \times 9 = (7 + 8) \times (7 + 9)$

We evaluate both sides following the order of operations.

Left side: $7 + 8 \times 9$. First, perform the multiplication.

$7 + (8 \times 9) = 7 + 72 = 79$.

Right side: $(7 + 8) \times (7 + 9)$. First, perform the additions in the parentheses.

$(7 + 8) = 15$.

$(7 + 9) = 16$.

Now, perform the multiplication: $15 \times 16$.

$\begin{array}{cc}& & 1 & 5 \\ \times & & 1 & 6 \\ \hline & & 9 & 0 \\ & 1 & 5 & \times \\ \hline 2 & 4 & 0 \\ \hline \end{array}$

So, $(7 + 8) \times (7 + 9) = 240$.

Since $79 \neq 240$, this statement is False.

(D) $7 \times (8 + 9) = (7 \times 8) + (7 \times 9)$

This is the distributive property of multiplication over addition.

Left side: $7 \times (8 + 9) = 7 \times 17$.

$\begin{array}{cc}& & 1 & 7 \\ \times & & & 7 \\ \hline & 1 & 1 & 9 \\ \hline \end{array}$

So, $7 \times (8 + 9) = 119$.

Right side: $(7 \times 8) + (7 \times 9) = 56 + 63$.

$\begin{array}{cc} & 5 & 6 \\ + & 6 & 3 \\ \hline 1 & 1 & 9 \\ \hline \end{array}$

So, $(7 \times 8) + (7 \times 9) = 119$.

Since $119 = 119$, this statement is True.

The statement which is not true is (C).

The correct answer is (C) 7 + 8 × 9 = (7 + 8) × (7 + 9).

We need to check which of the given numbers can be arranged in the form of a line, a triangle, and a rectangle using dot patterns.

Arrangement as a line:

Any whole number (greater than or equal to 1) can be arranged in a line of dots.

All the given options (9, 10, 11, 12) can be arranged in a line.

Arrangement as a triangle:

Numbers that can be arranged as a triangle are called triangular numbers. A triangular number is a number which is the sum of all integers from 1 up to some integer $n$. The $n$-th triangular number is given by the formula $T_n = \frac{n(n+1)}{2}$.

Let's check the options:

(A) 9: Is there an integer $n$ such that $\frac{n(n+1)}{2} = 9$? $n(n+1) = 18$. We check consecutive integers: $1 \times 2 = 2$, $2 \times 3 = 6$, $3 \times 4 = 12$, $4 \times 5 = 20$. No integer $n$ satisfies this. So, 9 is not a triangular number.

(B) 10: Is there an integer $n$ such that $\frac{n(n+1)}{2} = 10$? $n(n+1) = 20$. We see that $4 \times 5 = 20$. So, $n=4$. $T_4 = \frac{4(4+1)}{2} = \frac{4 \times 5}{2} = \frac{20}{2} = 10$. So, 10 is a triangular number.

(C) 11: Is there an integer $n$ such that $\frac{n(n+1)}{2} = 11$? $n(n+1) = 22$. We check consecutive integers: $4 \times 5 = 20$, $5 \times 6 = 30$. No integer $n$ satisfies this. So, 11 is not a triangular number.

(D) 12: Is there an integer $n$ such that $\frac{n(n+1)}{2} = 12$? $n(n+1) = 24$. We check consecutive integers: $4 \times 5 = 20$, $5 \times 6 = 30$. No integer $n$ satisfies this. So, 12 is not a triangular number.

Only 10 can be arranged as a triangle.

Arrangement as a rectangle:

Numbers that can be arranged as a rectangle are composite numbers (including perfect squares greater than 1). A number $N$ can form a rectangle if it can be written as the product of two integers $a$ and $b$, where $a > 1$ and $b > 1$.

Let's check the options:

(A) 9: $9 = 3 \times 3$. It can form a $3 \times 3$ square (a type of rectangle). So, 9 can be arranged as a rectangle.

(B) 10: $10 = 2 \times 5$. It can form a $2 \times 5$ rectangle. So, 10 can be arranged as a rectangle.

(C) 11: 11 is a prime number. Its only factors are 1 and 11. It cannot be expressed as $a \times b$ where $a > 1$ and $b > 1$. So, 11 cannot be arranged as a rectangle (other than a $1 \times 11$ line).

(D) 12: $12 = 2 \times 6$ or $12 = 3 \times 4$. It can form a $2 \times 6$ or $3 \times 4$ rectangle. So, 12 can be arranged as a rectangle.

Now let's see which number can be arranged in all three ways (line, triangle, and rectangle):

- 9: Line (Yes), Triangle (No), Rectangle (Yes). Fails for triangle.

- 10: Line (Yes), Triangle (Yes), Rectangle (Yes). Satisfies all three.

- 11: Line (Yes), Triangle (No), Rectangle (No). Fails for triangle and rectangle.

- 12: Line (Yes), Triangle (No), Rectangle (Yes). Fails for triangle.

The only number that can be arranged in all three ways is 10.

The correct answer is (B) 10.

We need to examine each statement about the properties of whole numbers ($W = \{0, 1, 2, 3, ...\}$) to find which one is not true.

(A) Both addition and multiplication are associative for whole numbers.

Associativity of Addition: For any whole numbers $a, b, c$, $(a + b) + c = a + (b + c)$. This is true for whole numbers. Example: $(2+3)+4 = 5+4 = 9$ and $2+(3+4) = 2+7 = 9$.

Associativity of Multiplication: For any whole numbers $a, b, c$, $(a \times b) \times c = a \times (b \times c)$. This is true for whole numbers. Example: $(2 \times 3) \times 4 = 6 \times 4 = 24$ and $2 \times (3 \times 4) = 2 \times 12 = 24$.

This statement is True.

(B) Zero is the identity for multiplication of whole numbers.

The identity element for an operation is the element that, when combined with any other element, leaves the other element unchanged.

For multiplication, the identity element $e$ must satisfy $a \times e = a$ and $e \times a = a$ for all whole numbers $a$.

Let's check if zero is the identity for multiplication:

Consider a non-zero whole number, say $a = 5$.

$5 \times 0 = 0$.

Since $5 \times 0 \neq 5$, $0$ is not the identity for multiplication.

The identity for multiplication of whole numbers is $1$ (since $a \times 1 = a$ and $1 \times a = a$ for any whole number $a$).

This statement is False.

(C) Addition and multiplication both are commutative for whole numbers.

Commutativity of Addition: For any whole numbers $a, b$, $a + b = b + a$. This is true for whole numbers. Example: $2 + 3 = 5$ and $3 + 2 = 5$.

Commutativity of Multiplication: For any whole numbers $a, b$, $a \times b = b \times a$. This is true for whole numbers. Example: $2 \times 3 = 6$ and $3 \times 2 = 6$.

This statement is True.

(D) Multiplication is distributive over addition for whole numbers.

Distributive Property: For any whole numbers $a, b, c$, $a \times (b + c) = (a \times b) + (a \times c)$. This is true for whole numbers. Example: $2 \times (3 + 4) = 2 \times 7 = 14$. And $(2 \times 3) + (2 \times 4) = 6 + 8 = 14$.

This statement is True.

The only statement that is not true is (B).

The correct answer is (B) Zero is the identity for muliplication of whole numbers.

We need to evaluate each given mathematical statement.

(A) $0 + 0 = 0$

The sum of $0$ and $0$ is $0$. This statement is True.

(B) $0 - 0 = 0$

The difference between $0$ and $0$ is $0$. This statement is True.

(C) $0 \times 0 = 0$

The product of $0$ and $0$ is $0$. This statement is True.

(D) $0 \div 0 = 0$

Division by zero is undefined. The expression $0 \div 0$ is an indeterminate form and is not equal to $0$.

If we assume $0 \div 0 = x$, then by the definition of division, $0 \times x = 0$. This equation is true for any value of $x$, which means $x$ is not uniquely determined. Thus, $0 \div 0$ is undefined or indeterminate.

Therefore, the statement $0 \div 0 = 0$ is False.

The statement which is not true is (D).

The correct answer is (D) 0 ÷ 0 = 0.

We need to find the predecessor of the number '1 lakh'.

In the Indian System of Numeration, 1 lakh is written as $1,00,000$.

The predecessor of a number is the number that comes immediately before it. It is found by subtracting $1$ from the number.

Predecessor of 1 lakh = $1,00,000 - 1$.

Let's perform the subtraction:

$\begin{array}{ccccccc} & 1 & 0 & 0 & 0 & 0 & 0 \\ - & & & & & & 1 \\ \hline & & 9 & 9 & 9 & 9 & 9 \\ \hline \end{array}$

So, the predecessor of $1,00,000$ is $99,999$.

Comparing this result with the given options:

(A) $99000$

(B) $99999$

(C) $999999$ (This is the greatest 6-digit number)

(D) $100001$ (This is the successor of 1 lakh)

The result $99999$ matches option (B).

The correct answer is (B) 99999.

We need to find the successor of 1 million.

In the International System of Numeration, 1 million is written as $1,000,000$.

The successor of a number is the number that comes immediately after it. It is obtained by adding $1$ to the number.

Successor of 1 million = $1,000,000 + 1$.

$1,000,000 + 1 = 1,000,001$.

Let's compare this result with the given options:

(A) 2 millions = $2,000,000$

(B) 1000001 = $1,000,001$

(C) 100001 = $100,001$

(D) 10001 = $10,001$

The result $1,000,001$ matches option (B).

The correct answer is (B) 1000001.

We need to find the number of even numbers that are strictly between $58$ and $80$.

The even numbers between $58$ and $80$ are those whole numbers $n$ such that $58 < n < 80$ and $n$ is divisible by $2$.

The first even number greater than $58$ is $60$.

The last even number less than $80$ is $78$.

The sequence of even numbers between $58$ and $80$ is $60, 62, 64, ..., 76, 78$.

To find the number of terms in this sequence, we can consider it as an arithmetic progression.

The first term is $a_1 = 60$.

The last term is $a_n = 78$.

The common difference is $d = 2$ (since they are consecutive even numbers).

The formula for the $n$-th term of an arithmetic progression is $a_n = a_1 + (n-1)d$.

Substituting the values:

$78 = 60 + (n-1) \times 2$

Subtract $60$ from both sides:

$78 - 60 = (n-1) \times 2$

$18 = (n-1) \times 2$

Divide by $2$:

$\frac{18}{2} = n-1$

$9 = n-1$

Add $1$ to both sides:

$n = 9 + 1 = 10$.

There are $10$ even numbers between $58$ and $80$.

Alternatively, we can find the number of even numbers up to $78$ and subtract the number of even numbers up to $58$.

The even numbers starting from 2 are $2 \times 1, 2 \times 2, ..., 2 \times k$. The number of even numbers up to $2k$ is $k$.

The number of even numbers up to $78$ is $\frac{78}{2} = 39$. (These are $2, 4, ..., 78$)

The number of even numbers up to $58$ is $\frac{58}{2} = 29$. (These are $2, 4, ..., 58$)

The even numbers between $58$ and $80$ are the even numbers from $60$ up to $78$.

Number of even numbers between $58$ and $80$ = (Number of even numbers $\leq 78$) - (Number of even numbers $\leq 58$).

This approach isn't quite right because it includes 58 and 80 if they were even. A better boundary is numbers $\leq 79$ and $\leq 59$.

The even numbers are $60, 62, ..., 78$.

Consider the list $2, 4, ..., 58, [60, 62, ..., 78], 80, 82, ...$

The numbers we want are in the bracket.

Number of even numbers from $2$ to $78$ is $78 \div 2 = 39$.

Number of even numbers from $2$ to $58$ is $58 \div 2 = 29$.

The even numbers greater than $58$ and less than or equal to $78$ are $39 - 29 = 10$.

Since $78$ is less than $80$, this count is correct for numbers between $58$ and $80$.

There are $10$ even numbers between $58$ and $80$.

Comparing the result with the given options:

(A) $10$

(B) $11$

(C) $12$

(D) $13$

The result $10$ matches option (A).

The correct answer is (A) 10.

We need to find the number of prime numbers in two intervals and then find their sum.

A prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.

Primes between 16 and 80:

We list the prime numbers greater than $16$ and less than $80$.

$17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79$.

Let's count them: There are $16$ prime numbers between $16$ and $80$.

Primes between 90 and 100:

We list the prime numbers greater than $90$ and less than $100$.

Check numbers from $91$ to $99$.

$91 = 7 \times 13$ (not prime)

$92$ (even, not prime)

$93 = 3 \times 31$ (not prime)

$94$ (even, not prime)

$95 = 5 \times 19$ (not prime)

$96$ (even, not prime)

$97$ (only divisors are 1 and 97, so it is prime)

$98$ (even, not prime)

$99 = 3 \times 33 = 9 \times 11$ (not prime)

The only prime number between $90$ and $100$ is $97$.

Let's count them: There is $1$ prime number between $90$ and $100$.

Sum of the number of primes:

Sum = (Number of primes between 16 and 80) + (Number of primes between 90 and 100)

Sum = $16 + 1 = 17$.

Comparing the result with the given options:

(A) $20$

(B) $18$

(C) $17$

(D) $16$

The result $17$ matches option (C).

The correct answer is (C) 17.

We will examine each statement to determine which one is not true.

(A) The HCF of two distinct prime numbers is 1.

A prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.

If we have two distinct prime numbers, say $p$ and $q$, their only positive divisors are $\{1, p\}$ and $\{1, q\}$ respectively. Since $p \neq q$, the only common positive divisor is $1$.

Therefore, the Highest Common Factor (HCF) of two distinct prime numbers is indeed $1$.

This statement is True.

(B) The HCF of two co prime numbers is 1.

Two numbers are said to be coprime (or relatively prime) if their only common positive divisor is $1$.

By definition, the HCF of two coprime numbers is $1$.

This statement is True.

(C) The HCF of two consecutive even numbers is 2.

Let the two consecutive even numbers be $2n$ and $2n+2$, where $n$ is a positive integer.

Both $2n$ and $2n+2$ are divisible by $2$. Thus, $2$ is a common factor.

Any common factor of $2n$ and $2n+2$ must also divide their difference: $(2n+2) - 2n = 2$.

The positive divisors of $2$ are $1$ and $2$.

Therefore, the greatest common factor can be at most $2$. Since $2$ is a common factor, the HCF must be $2$.

Example: HCF(4, 6) = 2, HCF(10, 12) = 2.

This statement is True.

(D) The HCF of an even and an odd number is even.

An even number is any integer that is divisible by $2$.

An odd number is any integer that is not divisible by $2$.

The HCF of two numbers is the greatest number that divides both of them.

If a number is even, it has a factor of $2$. If a number is odd, it does not have a factor of $2$.

For a number to be a common factor of an even and an odd number, it must divide both. Since an odd number is not divisible by $2$, a common factor cannot have $2$ as a factor.

Therefore, the HCF of an even number and an odd number cannot be even (as an even number is divisible by 2). The HCF must be an odd number.

Example: HCF(6, 9) = 3 (which is odd). HCF(12, 5) = 1 (which is odd).

This statement is False.

The statement which is not true is (D).

The correct answer is (D) The HCF of an even and an odd number is even.

The largest 4-digit number is $9999$.

We need to find the prime factors of $9999$.

We can perform prime factorization:

$\begin{array}{c|cc} 3 & 9999 \\ \hline 3 & 3333 \\ \hline 11 & 1111 \\ \hline 101 & 101 \\ \hline & 1 \end{array}$

The prime factorization of $9999$ is $3 \times 3 \times 11 \times 101$, which can be written as $3^2 \times 11^1 \times 101^1$.

The distinct prime factors are the unique prime numbers that divide the number.

From the prime factorization $3^2 \times 11^1 \times 101^1$, the distinct prime factors are $3$, $11$, and $101$.

The number of distinct prime factors is the count of these unique prime numbers.

There are $3$ distinct prime factors.

Comparing the result with the given options:

(A) $2$

(B) $3$

(C) $5$

(D) $11$

The number of distinct prime factors, $3$, matches option (B).

The correct answer is (B) 3.

The smallest 5-digit number is $10000$.

We need to find the distinct prime factors of $10000$.

We can perform prime factorization of $10000$.

$10000 = 10 \times 1000$

$10 = 2 \times 5$

$1000 = 10 \times 100 = (2 \times 5) \times 100$

$100 = 10 \times 10 = (2 \times 5) \times (2 \times 5)$

So, $10000 = 10 \times 10 \times 10 \times 10 = (2 \times 5) \times (2 \times 5) \times (2 \times 5) \times (2 \times 5)$.

The prime factorization is $2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5$.

In terms of powers, $10000 = 2^4 \times 5^4$.

Alternatively, using the prime factorization table:

$\begin{array}{c|cc} 2 & 10000 \\ \hline 2 & 5000 \\ \hline 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $10000 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = 2^4 \times 5^4$.

The distinct prime factors are the unique prime bases in the prime factorization.

The distinct prime factors of $10000$ are $2$ and $5$.

The number of distinct prime factors is the count of these unique prime numbers, which is $2$.

Comparing this result with the given options:

(A) $2$

(B) $4$

(C) $6$

(D) $8$

The number of distinct prime factors, $2$, matches option (A).

The correct answer is (A) 2.

A number is divisible by $22$ if it is divisible by both $2$ and $11$.

Let the given number be $N = 7254*98$, where $*$ represents a single digit.

Divisibility by 2:

A number is divisible by $2$ if its last digit is even.

The last digit of $N$ is $8$, which is an even number.

So, the number $7254*98$ is divisible by $2$ for any digit at $*$.

Divisibility by 11:

A number is divisible by $11$ if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) is either $0$ or a multiple of $11$.

Let the digit at $*$ be $x$. The number is $7254x98$.

Sum of digits at odd places (1st, 3rd, 5th, 7th from the right): $8 + x + 5 + 7 = 20 + x$.

Sum of digits at even places (2nd, 4th, 6th from the right): $9 + 4 + 2 = 15$.

Difference = (Sum of digits at odd places) - (Sum of digits at even places)

Difference = $(20 + x) - 15 = 5 + x$.

For the number to be divisible by $11$, the difference $5 + x$ must be a multiple of $11$ (including $0$).

Since $x$ is a single digit ($0 \leq x \leq 9$), the possible values for $5 + x$ are:

If $x=0$, $5+x = 5$.

If $x=1$, $5+x = 6$.

If $x=2$, $5+x = 7$.

...

If $x=6$, $5+x = 11$.

...

If $x=9$, $5+x = 14$.

The only multiple of $11$ in the range of possible values for $5+x$ ($5$ to $14$) is $11$.

So, we must have $5 + x = 11$.

Solving for $x$:

$x = 11 - 5$

$x = 6$.

Thus, the digit at $*$ must be $6$.

The number is $7254698$.

Comparing the digit $6$ with the given options:

(A) $1$

(B) $2$

(C) $6$

(D) $0$

The digit $6$ matches option (C).

The correct answer is (C) 6.

Let the first odd number in the pair be $2n + 1$, where $n$ is an integer.

The next consecutive odd number is obtained by adding $2$ to the first odd number.

The next consecutive odd number = $(2n + 1) + 2 = 2n + 3$.

We need to find the sum of this pair of consecutive odd numbers:

Sum = $(2n + 1) + (2n + 3)$

Sum = $2n + 1 + 2n + 3$

Sum = $2n + 2n + 1 + 3$

Sum = $4n + 4$.

We can factor out the common factor $4$ from the sum:

Sum = $4(n + 1)$.

Since $n$ is an integer, $n+1$ is also an integer.

The sum of any pair of consecutive odd numbers is always of the form $4 \times (\text{an integer})$.

This means the sum is always divisible by $4$.

Since the sum is $4(n+1)$, it is always divisible by $4$.

Is it always divisible by a number larger than 4?

Consider the possible values of $n+1$. As $n$ takes consecutive integer values, $n+1$ also takes consecutive integer values ($1, 2, 3, ...$ if we start with positive odd numbers like $1, 3, 5, ...$).

If $n=0$, the odd numbers are $1, 3$, sum is $4(0+1)=4$. Divisible by 4.

If $n=1$, the odd numbers are $3, 5$, sum is $4(1+1)=8$. Divisible by 4 and 8.

If $n=2$, the odd numbers are $5, 7$, sum is $4(2+1)=12$. Divisible by 4, 6, 12.

If $n=3$, the odd numbers are $7, 9$, sum is $4(3+1)=16$. Divisible by 4, 8, 16.

The number that always divides the sum $4(n+1)$ for any integer $n$ is the factors of the smallest possible non-zero sum, and also a common factor of all possible sums.

The smallest positive sum occurs when $n=0$, giving $4$. The factors are $1, 2, 4$.

The next sum is $8$. The factors are $1, 2, 4, 8$.

The next sum is $12$. The factors are $1, 2, 3, 4, 6, 12$.

The common factors of $4, 8, 12, 16, ...$ are $1, 2, 4$.

The largest number that always divides the sum $4(n+1)$ is $4$.

Comparing this result with the given options:

(A) $2$ (Divides all sums, but not the largest)

(B) $4$ (Divides all sums)

(C) $6$ (Does not divide 4 or 16)

(D) $8$ (Does not divide 4 or 12)

The largest number that always divides the sum is $4$.

The correct answer is (B) 4.

If a number is divisible by two numbers, it is also divisible by their Least Common Multiple (LCM).

We are given that a number is divisible by $5$ and $6$.

Let's find the LCM of $5$ and $6$.

The prime factorization of $5$ is $5$.

The prime factorization of $6$ is $2 \times 3$.

Since $5$ and $6$ have no common prime factors (they are coprime), their LCM is the product of the numbers.

LCM$(5, 6) = 5 \times 6 = 30$.

This means that any number divisible by both $5$ and $6$ must also be divisible by $30$.

Now, let's check the given options:

(A) $10$: If a number is divisible by $30$, it is also divisible by $10$ because $30$ is a multiple of $10$ ($30 = 3 \times 10$). So, if a number is divisible by $5$ and $6$, it must be divisible by $10$.

(B) $15$: If a number is divisible by $30$, it is also divisible by $15$ because $30$ is a multiple of $15$ ($30 = 2 \times 15$). So, if a number is divisible by $5$ and $6$, it must be divisible by $15$.

(C) $30$: As concluded from the LCM, if a number is divisible by $5$ and $6$, it must be divisible by $30$.

(D) $60$: If a number is divisible by $30$, is it necessarily divisible by $60$? Not always. For example, the number $30$ itself is divisible by $5$ ($30 \div 5 = 6$) and divisible by $6$ ($30 \div 6 = 5$). However, $30$ is not divisible by $60$ ($30 \div 60$ is not a whole number).

Therefore, a number divisible by $5$ and $6$ (i.e., divisible by $30$) may or may not be divisible by $60$.

The correct answer is (D) 60.

We need to find the sum of the prime factors of the number $1729$.

First, let's find the prime factorization of $1729$.

We test divisibility by small prime numbers:

$1729$ is not divisible by $2$ (it's an odd number).

The sum of the digits is $1+7+2+9 = 19$. Since $19$ is not divisible by $3$, $1729$ is not divisible by $3$.

$1729$ does not end in $0$ or $5$, so it is not divisible by $5$.

Let's check divisibility by $7$:

$1729 \div 7$.

$17 \div 7 = 2$ remainder $3$.

$32 \div 7 = 4$ remainder $4$.

$49 \div 7 = 7$ remainder $0$.

So, $1729 = 7 \times 247$.

Now we need to factorize $247$. Let's continue testing prime numbers.

Is $247$ divisible by $7$? $247 \div 7$. $24 \div 7 = 3$ remainder $3$. $37 \div 7 = 5$ remainder $2$. No.

Is $247$ divisible by $11$? $247 \div 11$. $24 \div 11 = 2$ remainder $2$. $27 \div 11 = 2$ remainder $5$. No.

Is $247$ divisible by $13$? $247 \div 13$. $24 \div 13 = 1$ remainder $11$. $117 \div 13$. We know $13 \times 9 = 117$.

So, $247 = 13 \times 19$.

Both $13$ and $19$ are prime numbers.

The prime factorization of $1729$ is $7 \times 13 \times 19$.

The distinct prime factors of $1729$ are $7$, $13$, and $19$.

The sum of these distinct prime factors is $7 + 13 + 19$.

$7 + 13 + 19 = 20 + 19 = 39$.

Comparing this result with the given options:

(A) $13$

(B) $19$

(C) $32$

(D) $39$

The sum $39$ matches option (D).

The correct answer is (D) 39.

Let the odd natural number be $n$.

Since the number is an odd natural number other than $1$, the possible values for $n$ are $3, 5, 7, 9, ...$.

The predecessor of $n$ is $n-1$.

The successor of $n$ is $n+1$.

We are interested in the product of the predecessor and successor, which is $(n-1) \times (n+1)$.

Since $n$ is an odd natural number greater than $1$, we can represent $n$ as $2k+1$ for some integer $k \geq 1$.

The predecessor is $n-1 = (2k+1) - 1 = 2k$.

The successor is $n+1 = (2k+1) + 1 = 2k+2$.

The product is $(n-1)(n+1) = (2k)(2k+2)$.

Product $= 2k \times 2(k+1)$

Product $= 4k(k+1)$.

We know that the product of two consecutive integers, $k$ and $k+1$, is always an even number.

So, $k(k+1)$ is divisible by $2$. We can write $k(k+1) = 2m$ for some integer $m$.

Substituting this back into the product:

Product $= 4 \times (2m) = 8m$.

Since the product can always be expressed as $8m$ where $m$ is an integer (for $k \geq 1$), the product of the predecessor and successor of an odd natural number other than 1 is always divisible by $8$.

Let's verify with examples of odd natural numbers other than 1:

If $n=3$: Predecessor = 2, Successor = 4. Product = $2 \times 4 = 8$. $8$ is divisible by $8$. ($k=1$)

If $n=5$: Predecessor = 4, Successor = 6. Product = $4 \times 6 = 24$. $24$ is divisible by $8$ ($24 \div 8 = 3$). ($k=2$)

If $n=7$: Predecessor = 6, Successor = 8. Product = $6 \times 8 = 48$. $48$ is divisible by $8$ ($48 \div 8 = 6$). ($k=3$)

If $n=9$: Predecessor = 8, Successor = 10. Product = $8 \times 10 = 80$. $80$ is divisible by $8$ ($80 \div 8 = 10$). ($k=4$)

The values obtained ($8, 24, 48, 80, ...$) are all multiples of $8$.

The greatest number that always divides all these values is $8$.

Comparing this result with the given options:

(A) $6$

(B) $4$

(C) $16$

(D) $8$

The number $8$ matches option (D).

The correct answer is (D) 8.

We need to find the common prime factors of $75$, $60$, and $105$.

First, let's find the prime factorization of each number.

Prime factorization of $75$:

$75 = 3 \times 25 = 3 \times 5 \times 5 = 3^1 \times 5^2$.

Prime factors of $75$ are $3$ and $5$.

Prime factorization of $60$:

$60 = 6 \times 10 = (2 \times 3) \times (2 \times 5) = 2 \times 2 \times 3 \times 5 = 2^2 \times 3^1 \times 5^1$.

Prime factors of $60$ are $2, 3,$ and $5$.

Prime factorization of $105$:

$105 = 5 \times 21 = 5 \times (3 \times 7) = 3 \times 5 \times 7 = 3^1 \times 5^1 \times 7^1$.

Prime factors of $105$ are $3, 5,$ and $7$.

Now, we find the common prime factors that appear in the factorization of all three numbers ($75, 60, 105$).

Prime factors of $75$: $\{3, 5\}$

Prime factors of $60$: $\{2, 3, 5\}$

Prime factors of $105$: $\{3, 5, 7\}$

The common prime factors are the elements present in all three sets: $\{3, 5\}$.

The common prime factors are $3$ and $5$.

The number of common prime factors is the count of these distinct common prime numbers, which is $2$.

Note that this is equivalent to finding the number of distinct prime factors of the HCF of the three numbers.

HCF$(75, 60, 105)$.

$75 = 3^1 \times 5^2$

$60 = 2^2 \times 3^1 \times 5^1$

$105 = 3^1 \times 5^1 \times 7^1$

The common prime factors with the lowest powers are $3^1$ and $5^1$.

HCF$(75, 60, 105) = 3^1 \times 5^1 = 15$.

The prime factors of $15$ are $3$ and $5$.

The number of distinct prime factors of HCF is $2$.

Comparing the result with the given options:

(A) $2$

(B) $3$

(C) $4$

(D) $5$

The number of common prime factors, $2$, matches option (A).

The correct answer is (A) 2.

Two numbers are coprime (or relatively prime) if their Highest Common Factor (HCF) is $1$. We need to find the pair whose HCF is not $1$.

(A) 8, 10

Factors of 8: $1, 2, 4, 8$.

Factors of 10: $1, 2, 5, 10$.

The common factors are $1$ and $2$.

The HCF(8, 10) = 2.

Since the HCF is $2$ (which is not $1$), the pair $(8, 10)$ is not coprime.

Let's check the other options to confirm.

(B) 11, 12

Factors of 11: $1, 11$ (11 is a prime number).

Factors of 12: $1, 2, 3, 4, 6, 12$.

The only common factor is $1$.

The HCF(11, 12) = 1.

The pair $(11, 12)$ is coprime.

(C) 1, 3

Factors of 1: $1$.

Factors of 3: $1, 3$.

The only common factor is $1$.

The HCF(1, 3) = 1.

The pair $(1, 3)$ is coprime. (Note: 1 is coprime to every integer).

(D) 31, 33

31 is a prime number. Its factors are $1, 31$.

Factors of 33: $1, 3, 11, 33$.

The only common factor is $1$.

The HCF(31, 33) = 1.

The pair $(31, 33)$ is coprime.

The pair that is not coprime is (A) 8, 10.

The correct answer is (A) 8, 10.

A number is divisible by $11$ if the difference between the sum of its digits at odd places (from the right) and the sum of its digits at even places (from the right) is either $0$ or a multiple of $11$.

Let's check each option using the divisibility rule for $11$.

(A) 1011011

The digits from right to left are $1, 1, 0, 1, 1, 0, 1$.

Sum of digits at odd places (1st, 3rd, 5th, 7th) = $1 + 0 + 1 + 1 = 3$.

Sum of digits at even places (2nd, 4th, 6th) = $1 + 1 + 0 = 2$.

Difference = Sum(Odd Places) - Sum(Even Places) = $3 - 2 = 1$.

Since $1$ is not $0$ or a multiple of $11$, the number $1011011$ is not divisible by $11$.

(B) 1111111

The digits from right to left are $1, 1, 1, 1, 1, 1, 1$.

Sum of digits at odd places (1st, 3rd, 5th, 7th) = $1 + 1 + 1 + 1 = 4$.

Sum of digits at even places (2nd, 4th, 6th) = $1 + 1 + 1 = 3$.

Difference = $4 - 3 = 1$.

Since $1$ is not $0$ or a multiple of $11$, the number $1111111$ is not divisible by $11$.

(C) 22222222

The digits from right to left are $2, 2, 2, 2, 2, 2, 2, 2$.

Sum of digits at odd places (1st, 3rd, 5th, 7th) = $2 + 2 + 2 + 2 = 8$.

Sum of digits at even places (2nd, 4th, 6th, 8th) = $2 + 2 + 2 + 2 = 8$.

Difference = $8 - 8 = 0$.

Since $0$ is a multiple of $11$, the number $22222222$ is divisible by $11$.

(D) 3333333

The digits from right to left are $3, 3, 3, 3, 3, 3, 3$.

Sum of digits at odd places (1st, 3rd, 5th, 7th) = $3 + 3 + 3 + 3 = 12$.

Sum of digits at even places (2nd, 4th, 6th) = $3 + 3 + 3 = 9$.

Difference = $12 - 9 = 3$.

Since $3$ is not $0$ or a multiple of $11$, the number $3333333$ is not divisible by $11$.

The only number divisible by $11$ among the options is $22222222$.

The correct answer is (C) 22222222.

We need to find the Least Common Multiple (LCM) of $10, 15,$ and $20$.

We can use the prime factorization method or the common division method.

Method 1: Prime Factorization

Prime factorization of 10: $10 = 2 \times 5 = 2^1 \times 5^1$.

Prime factorization of 15: $15 = 3 \times 5 = 3^1 \times 5^1$.

Prime factorization of 20: $20 = 2 \times 10 = 2 \times 2 \times 5 = 2^2 \times 5^1$.

The LCM is found by taking the highest power of all the prime factors involved in the numbers. The prime factors involved are $2, 3,$ and $5$.

Highest power of 2 is $2^2$.

Highest power of 3 is $3^1$.

Highest power of 5 is $5^1$.

LCM$(10, 15, 20) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 12 \times 5 = 60$.

Method 2: Common Division Method

$\begin{array}{c|cc} 2 & 10 \;, & 15 \;, & 20 \\ \hline 2 & 5 \; , & 15 \; , & 10 \\ \hline 3 & 5 \; , & 15 \; , & 5 \\ \hline 5 & 5 \; , & 5 \; , & 5 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

The LCM is the product of the divisors: $2 \times 2 \times 3 \times 5 = 4 \times 15 = 60$.

Both methods give the same result. The LCM of $10, 15,$ and $20$ is $60$.

Comparing this result with the given options:

(A) $30$ (Divisible by 10 and 15, but not 20)

(B) $60$ (Divisible by 10, 15, and 20)

(C) $90$ (Divisible by 10 and 15, but not 20)

(D) $180$ (Divisible by 10, 15, and 20, but not the least common multiple)

The LCM $60$ matches option (B).

The correct answer is (B) 60.

We are given that the Least Common Multiple (LCM) of two numbers is $180$.

A fundamental property relating the Highest Common Factor (HCF) and LCM of two numbers is that the HCF of the numbers must always be a factor of their LCM.

In other words, if HCF = $h$ and LCM = $l$ for two numbers, then $l$ must be divisible by $h$.

Given LCM = $180$. So, the HCF of the two numbers must be a factor of $180$.

We need to check which of the given options is NOT a factor of $180$.

Let's check each option:

(A) 45: Is $180$ divisible by $45$? $180 \div 45 = 4$. Yes, $45$ is a factor of $180$.

(B) 60: Is $180$ divisible by $60$? $180 \div 60 = 3$. Yes, $60$ is a factor of $180$.

(C) 75: Is $180$ divisible by $75$? $180 \div 75 = \frac{180}{75} = \frac{\cancel{180}^{36}}{\cancel{75}_{15}} = \frac{36}{15} = \frac{\cancel{36}^{12}}{\cancel{15}_{5}} = \frac{12}{5} = 2.4$. This is not a whole number. So, $75$ is not a factor of $180$.

(D) 90: Is $180$ divisible by $90$? $180 \div 90 = 2$. Yes, $90$ is a factor of $180$.

Since the HCF of two numbers must be a factor of their LCM, and $75$ is not a factor of $180$, $75$ cannot be the HCF of two numbers whose LCM is $180$.

The correct answer is (C) 75.

The statement is True.

In Roman numeration, the symbols I, X, C, and M can be repeated up to three times consecutively to denote a sum (e.g., III = 3, XX = 20, CCC = 300, MMM = 3000). Symbols V, L, and D are never repeated.

The statement is False.

In Roman numeration, when a symbol is repeated, its value is added as many times as it occurs, not multiplied.

For example:

II means $1 + 1 = 2$ (not $1 \times 1 = 1$)

XX means $10 + 10 = 20$ (not $10 \times 10 = 100$)

The statement is True.

The expression $5 \times 1000 + 5 \times 100 + 5 \times 10 + 5 \times 1$ is the expanded form of the number 5555 based on its place values.

Calculation:

$5 \times 1000 = 5000$

$5 \times 100 = 500$

$5 \times 10 = 50$

$5 \times 1 = 5$

Adding these values:

$5000 + 500 + 50 + 5 = 5555$

So, $5555 = 5555$, which is correct.

The statement is True.

The given expression is the expanded form of the number 39746 based on its place values.

Let's calculate the value of the right side:

$3 \times 10000 = 30000$

$9 \times 1000 = 9000$

$7 \times 100 = 700$

$4 \times 10 = 40$

$6$

Summing these values:

$30000 + 9000 + 700 + 40 + 6 = 39746$

Since $39746 = 39746$, the statement is correct.

The statement is False.

The expanded form of 82546 should be based on the place value of each digit.

The place values are:

• 8 is in the ten thousands place ($10000$)
• 2 is in the thousands place ($1000$)
• 5 is in the hundreds place ($100$)
• 4 is in the tens place ($10$)
• 6 is in the ones place ($1$)

So, the correct expanded form is:

$8 \times 10000 + 2 \times 1000 + 5 \times 100 + 4 \times 10 + 6 \times 1$

Let's calculate the value of the expression given in the statement:

$8 \times 1000 = 8000$

$2 \times 1000 = 2000$

$5 \times 100 = 500$

$4 \times 10 = 40$

$6$

Adding these values:

$8000 + 2000 + 500 + 40 + 6 = 10546$

Since $82546 \neq 10546$, the given statement is incorrect.

The statement is True.

The given expression is the expanded form of the number 532235 based on the place value of each digit.

Let's calculate the value of the right side:

$5 \times 100000 = 500000$

$3 \times 10000 = 30000$

$2 \times 1000 = 2000$

$2 \times 100 = 200$

$3 \times 10 = 30$

$5 \times 1 = 5$

Summing these values:

$500000 + 30000 + 2000 + 200 + 30 + 5 = 532235$

Since $532235 = 532235$, the statement is correct.

The statement is False.

In Roman numeration:

• XX represents $10 + 10 = 20$.
• IX represents $10 - 1 = 9$ (since the symbol I, representing a smaller value, is placed before X, representing a larger value).

Combining these values:

XXIX = XX + IX = $20 + 9 = 29$.

Therefore, XXIX is equal to 29, not 31.

The statement is True.

In Roman numeration:

• L represents 50.
• XX represents $10 + 10 = 20$.
• IV represents $5 - 1 = 4$ (due to the subtraction rule, where a smaller value I is placed before a larger value V).

Combining these values:

LXXIV = L + XX + IV = $50 + 20 + 4 = 74$.

Therefore, LXXIV is indeed equal to 74.

The statement is False.

Let's convert the Roman numerals to Hindu-Arabic numerals:

• LIV: L represents 50, IV represents $5 - 1 = 4$. So, LIV = $50 + 4 = 54$.
• LVI: L represents 50, V represents 5, I represents 1. So, LVI = $50 + 5 + 1 = 56$.

Comparing the values:

LIV = 54

LVI = 56

Since $54 < 56$, LIV is not greater than LVI. LVI is greater than LIV.

The statement is False.

Descending order means arranging numbers from the largest to the smallest.

The given numbers are: 4578, 4587, 5478, 5487.

Let's compare them:

• Comparing 4578 and 4587: 4587 > 4578
• Comparing 5478 and 5487: 5487 > 5478
• Comparing 4587 and 5478: 5478 > 4587 (comparing the thousands digit)

Arranging the numbers in descending order:

5487, 5478, 4587, 4578

The given order is 4578, 4587, 5478, 5487, which is actually in ascending order (from smallest to largest).

The statement is False.

To round off 85764 to the nearest hundreds, we look at the tens digit, which is 6.

Since the tens digit (6) is 5 or greater, we round up the hundreds digit.

The hundreds digit is 7. Rounding up 7 gives 8.

The digits to the right of the hundreds place become zeros.

So, 85764 rounded off to the nearest hundreds is 85800.

The statement is False.

First, let's round off each number to the nearest hundreds:

• 7826: The tens digit is 2. Since $2 < 5$, we round down the hundreds digit. 7826 rounded to the nearest hundreds is 7800.
• 12469: The tens digit is 6. Since $6 \geq 5$, we round up the hundreds digit. 12469 rounded to the nearest hundreds is 12500.

Now, let's find the estimated sum by adding the rounded numbers:

$7800 + 12500 = 20300$

The estimated sum rounded off to hundreds is 20300, not 20000.

The statement is False.

To form the largest possible six-digit number using the digits 5, 3, 4, 7, 0, and 8 exactly once, we must arrange the digits in descending order from left to right (from the largest place value to the smallest).

The given digits are 8, 7, 5, 4, 3, 0.

Arranging them in descending order gives: 8, 7, 5, 4, 3, 0.

The largest number formed by these digits is 875430.

The given number is 875403.

Comparing the two numbers, $875430 > 875403$.

Therefore, the largest six-digit number is 875430, not 875403.

The statement is False.

Let's read the number 81652318 using the Indian place value system. We group the digits from right to left in periods of 3, 2, 2, and so on.

81,65,23,318

Reading by periods:

• Crores period: 81 (Eighty-one crore)
• Lakhs period: 65 (Sixty-five lakh)
• Thousands period: 23 (Twenty-three thousand)
• Ones period: 318 (Three hundred eighteen)

Putting it together, the number 81652318 is read as Eighty-one crore, sixty-five lakh, twenty-three thousand, three hundred eighteen.

The given reading is "eighty one crore six lakh fifty two thousand three hundred eighteen", which is incorrect in the lakhs period (it should be sixty-five lakh) and the thousands period (it should be twenty-three thousand). It seems the number intended might have been 8,16,52,318 which would be read as eighty-one lakh, sixty-five thousand, two hundred thirty-one (or read as eighty-one lakh fifty-two thousand three hundred eighteen based on spacing). However, based on the exact digits 81652318, the reading is as derived above.

Assuming the number is 81,652,318, the correct reading in the Indian system is Eighty-one crore, sixty-five lakh, twenty-three thousand, three hundred eighteen.

The statement is True.

To form the largest 4-digit number using the digits 6, 7, 0, and 9 exactly once, we need to arrange the digits in descending order.

The given digits are 9, 7, 6, 0.

Arranging them in descending order gives: 9, 7, 6, 0.

Placing these digits in the thousands, hundreds, tens, and ones places respectively, we get the number 9760.

This is the largest 4-digit number that can be formed using these digits.

The statement is False.

These are prefixes used in the metric system, representing powers of 10.

Let's look at their values relative to the base unit (e.g., meter, gram, liter):

• Kilo (k) means $10^3 = 1000$ times the base unit.
• Centi (c) means $10^{-2} = \frac{1}{100} = 0.01$ times the base unit.
• Milli (m) means $10^{-3} = \frac{1}{1000} = 0.001$ times the base unit.

Comparing the values:

• Kilo is the largest (1000).
• Centi is smaller (0.01).
• Milli is the smallest (0.001).

Therefore, among kilo, milli, and centi, the smallest is milli.

The statement is False.

The successor of a number is the number that comes immediately after it, which is obtained by adding 1 to the number.

One-digit numbers are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

Let's find the successor of the largest one-digit number, which is 9.

Successor of 9 = $9 + 1 = 10$.

10 is a two-digit number.

Since the successor of 9 is 10, which is not a one-digit number, the statement is false.

The statement is False.

The successor of a number is obtained by adding 1 to it.

Consider the largest 3-digit number, which is 999.

The successor of 999 is $999 + 1 = 1000$.

1000 is a 4-digit number.

Since the successor of the largest 3-digit number (999) is a 4-digit number (1000), the statement is false.

The statement is False.

The predecessor of a number is the number that comes immediately before it, which is obtained by subtracting 1 from the number.

Two-digit numbers range from 10 to 99.

Let's find the predecessor of the smallest two-digit number, which is 10.

Predecessor of 10 = $10 - 1 = 9$.

9 is a one-digit number.

Since the predecessor of 10 is 9, which is not a two-digit number, the statement is false.

The statement is True.

Whole numbers are $0, 1, 2, 3, 4, \dots$ (non-negative integers).

The successor of any whole number $n$ is $n + 1$. Since the set of whole numbers is infinite and for any whole number $n$, $n+1$ is also a whole number, every whole number has a unique successor which is also a whole number.

There is no largest whole number.

The statement is False.

Whole numbers are $0, 1, 2, 3, 4, \dots$

The predecessor of a number is obtained by subtracting 1 from it.

Let's find the predecessor of the smallest whole number, which is 0.

Predecessor of 0 = $0 - 1 = -1$.

-1 is an integer, but it is not a whole number (since whole numbers are non-negative).

Therefore, the whole number 0 does not have a predecessor within the set of whole numbers.

The statement is False.

Natural numbers are $1, 2, 3, 4, \dots$ (positive integers).

Consider two consecutive natural numbers, for example, 1 and 2.

There is no natural number between 1 and 2.

If we take two non-consecutive natural numbers, say 3 and 5, there is one natural number (4) between them. But this is not true for *any* two natural numbers.

For the statement to be true, there must always be at least one natural number between any two given natural numbers. As shown with 1 and 2, this is not the case.

The statement is True.

The largest 3-digit number is 999.

The smallest 4-digit number is 1000.

The successor of the largest 3-digit number (999) is obtained by adding 1 to it:

Successor of 999 = $999 + 1 = 1000$.

Since the successor of the largest 3-digit number (1000) is equal to the smallest 4-digit number (1000), the statement is true.

The statement is True.

This is a fundamental rule when comparing natural numbers (or positive integers).

A number with more digits will always have a value greater than any number with fewer digits, regardless of the value of the digits themselves.

For example:

• The smallest 2-digit number is 10. The largest 1-digit number is 9. $10 > 9$.
• The smallest 3-digit number is 100. The largest 2-digit number is 99. $100 > 99$.
• Any 5-digit number (like 10000) is greater than any 4-digit number (like 9999).

This is because the leftmost digit in the number with more digits represents a much larger place value than any digit in the number with fewer digits.

The statement is True.

A set of numbers is closed under an operation if performing that operation on any two numbers in the set always results in a number that is also in the same set.

Natural numbers are $1, 2, 3, 4, \dots$.

Addition is the operation.

If we take any two natural numbers, say $a$ and $b$, and add them ($a + b$), the result will always be a natural number.

For example:

• $1 + 1 = 2$ (2 is a natural number)
• $5 + 8 = 13$ (13 is a natural number)
• $100 + 200 = 300$ (300 is a natural number)

There is no pair of natural numbers whose sum is not a natural number.

Therefore, natural numbers are closed under addition.

The statement is False.

Natural numbers are $1, 2, 3, 4, \dots$.

A set is closed under an operation if the result of the operation on any two elements of the set is also in the set.

Multiplication is the operation.

If we take any two natural numbers, say $a$ and $b$, and multiply them ($a \times b$), the result will always be a natural number.

For example:

• $1 \times 1 = 1$ (1 is a natural number)
• $2 \times 3 = 6$ (6 is a natural number)
• $10 \times 5 = 50$ (50 is a natural number)

There is no pair of natural numbers whose product is not a natural number.

Therefore, natural numbers are closed under multiplication.

The statement is False.

Natural numbers are $1, 2, 3, 4, \dots$.

A set is closed under an operation if the result of the operation on any two elements of the set is also in the set.

Subtraction is the operation.

If we take two natural numbers, say $a$ and $b$, and subtract them ($a - b$), the result is not always a natural number.

For example:

• $2 - 3 = -1$. -1 is an integer but not a natural number.
• $5 - 5 = 0$. 0 is a whole number but not a natural number (if natural numbers start from 1).
• $10 - 12 = -2$. -2 is not a natural number.

Since the result of subtracting two natural numbers can be an integer that is not a natural number, the set of natural numbers is not closed under subtraction.

The statement is True.

An operation is commutative for a set of numbers if the order of the operands does not affect the result.

For addition, this means that for any two natural numbers $a$ and $b$, $a + b = b + a$.

Natural numbers are $1, 2, 3, 4, \dots$.

Let's test with examples:

• $2 + 3 = 5$ and $3 + 2 = 5$. So, $2 + 3 = 3 + 2$.
• $10 + 7 = 17$ and $7 + 10 = 17$. So, $10 + 7 = 7 + 10$.
• For any natural numbers $a$ and $b$, their sum is the same regardless of the order.

Therefore, addition is commutative for natural numbers.

The statement is False.

An additive identity is a number such that when it is added to any number in a set, the result is the original number.

For whole numbers (0, 1, 2, 3, ...), the additive identity is 0.

This is because for any whole number $a$, $a + 0 = a$ and $0 + a = a$.

For example:

• $5 + 0 = 5$
• $0 + 12 = 12$

If we use 1 as the identity:

$5 + 1 = 6$, which is not equal to 5.

Therefore, 1 is not the additive identity for whole numbers. 0 is the additive identity.

The statement is True.

A multiplicative identity is a number such that when it is multiplied by any number in a set, the result is the original number.

For whole numbers (0, 1, 2, 3, ...), the multiplicative identity is 1.

This is because for any whole number $a$, $a \times 1 = a$ and $1 \times a = a$.

For example:

• $5 \times 1 = 5$
• $1 \times 12 = 12$
• $0 \times 1 = 0$

Therefore, 1 is the multiplicative identity for whole numbers.

The statement is True.

The whole number described in the statement is the additive identity for whole numbers.

The whole number is 0.

When 0 is added to any whole number $a$, the result is $a$ itself ($a + 0 = a$).

For example:

• $5 + 0 = 5$
• $0 + 0 = 0$
• $100 + 0 = 100$

So, there exists such a whole number, which is 0.

The statement is False.

We are looking for a natural number, let's call it $e$, such that for any natural number $a$, $a + e = a$.

If $a + e = a$, then subtracting $a$ from both sides gives $e = a - a = 0$.

The number required for this property is 0.

However, natural numbers are $1, 2, 3, \dots$. The number 0 is not a natural number.

Therefore, there is no natural number which, when added to a natural number, gives the number itself.

Given:

A whole number is divided by another whole number which is greater than the first one.

To Check:

The truth value of the statement: "the quotient is not equal to zero".

Solution:

Let the first whole number be $a$ and the second whole number be $b$.

According to the problem statement, $a$ and $b$ are whole numbers.

This means $a \in \{0, 1, 2, 3, ...\}$ and $b \in \{0, 1, 2, 3, ...\}$.

We are also given that the second number is greater than the first one, so $b > a$.

Since $b > a$ and $a$ is a whole number ($a \ge 0$), $b$ must be a positive whole number. Specifically, if $a=0$, then $b>0$; if $a>0$, then $b>a \ge 1$. In any case, $b \ne 0$.

We are considering the quotient of dividing $a$ by $b$, which is $\frac{a}{b}$.

Let's examine the statement by considering a specific case.

Consider the case where the first whole number $a = 0$.

According to the given condition, the second whole number $b$ must be greater than the first one, i.e., $b > 0$.

Since $b$ is a whole number and $b > 0$, $b$ can be any positive integer such as $1, 2, 3, ...$.

Let's take $b = 5$. Both $a=0$ and $b=5$ are whole numbers, and $b > a$ ($5 > 0$).

Now, we find the quotient when $a$ is divided by $b$, which is $\frac{a}{b} = \frac{0}{5}$.

The value of this quotient is calculated as follows:

In this specific case, where a whole number ($0$) is divided by another whole number which is greater than the first one ($5$), the quotient is found to be equal to zero.

The original statement claims that for any such division, the quotient is not equal to zero.

Since we have found a valid scenario (dividing $0$ by a larger whole number) where the quotient is equal to zero, the statement is contradicted.

Therefore, the given statement is false.

Answer:

The statement is False.

Given:

A non-zero whole number is divided by itself.

To Check:

The truth value of the statement: "the quotient is 1".

Solution:

Let the non-zero whole number be $a$.

According to the problem statement, $a$ is a whole number and $a \neq 0$.

This means $a$ belongs to the set of positive integers, i.e., $a \in \{1, 2, 3, ...\}$.

We are considering the division of this number by itself, which can be represented as $a \div a$ or $\frac{a}{a}$.

For any non-zero number $a$, the division of $a$ by itself is defined as the unique number $q$ such that $q \times a = a$. The only number $q$ that satisfies this equation for any $a \neq 0$ is $1$.

Mathematically, for any $a \ne 0$, the quotient is:

$\frac{a}{a} = 1$

This property holds true for all non-zero whole numbers.

For instance, if the non-zero whole number is $7$, dividing it by itself gives $\frac{7}{7} = 1$.

If the non-zero whole number is $42$, dividing it by itself gives $\frac{42}{42} = 1$.

Since the result of dividing any non-zero whole number by itself is always $1$, the given statement is correct.

Answer:

The statement is True.

Given:

The product of two whole numbers.

To Check:

The truth value of the statement: "the product need not be a whole number".

Solution:

Whole numbers are the set of non-negative integers: $\{0, 1, 2, 3, ...\}$.

Let $a$ and $b$ be any two whole numbers.

We need to consider their product, which is $a \times b$ or $ab$.

One of the fundamental properties of the set of whole numbers under the operation of multiplication is called closure.

The property of closure states that if you perform the operation on any two elements within the set, the result is also an element within the same set.

For multiplication of whole numbers, the closure property means that for any two whole numbers $a$ and $b$, their product $a \times b$ is always a whole number.

Let's look at a few examples:

If $a=3$ and $b=5$, their product is $3 \times 5 = 15$. $15$ is a whole number.

If $a=0$ and $b=7$, their product is $0 \times 7 = 0$. $0$ is a whole number.

If $a=10$ and $b=12$, their product is $10 \times 12 = 120$. $120$ is a whole number.

In every case where we multiply two whole numbers, the result is invariably a whole number.

Therefore, the product of two whole numbers is always a whole number.

The statement "The product of two whole numbers need not be a whole number" is equivalent to saying that sometimes the product of two whole numbers is not a whole number. This contradicts the closure property of whole numbers under multiplication.

Hence, the given statement is false.

Answer:

The statement is False.

Given:

A whole number is divided by another whole number greater than 1.

To Check:

The truth value of the statement: "the quotient is never equal to the former number".

Solution:

Let the first whole number be $a$ and the second whole number be $b$.

According to the problem, $a$ and $b$ are whole numbers, so $a \in \{0, 1, 2, 3, ...\}$ and $b \in \{0, 1, 2, 3, ...\}$.

We are also given that the second whole number is greater than 1, which means $b \in \{2, 3, 4, ...\}$. Note that since $b > 1$, $b$ is a positive whole number, so $b \neq 0$.

We are considering the quotient of dividing $a$ by $b$, which is $\frac{a}{b}$.

The statement claims that this quotient is never equal to the former number $a$. In other words, for all whole numbers $a$ and whole numbers $b > 1$, the statement claims that $\frac{a}{b} \neq a$.

Let's investigate if there is any case where the quotient $\frac{a}{b}$ is equal to $a$. We can write this as an equation:

We can solve this equation for $a$, keeping in mind the conditions on $a$ and $b$. Since $b > 1$, $b$ is not zero, so we can multiply both sides by $b$:

Now, rearrange the equation to one side:

Factor out $a$ from the left side:

For the product of two numbers to be zero, at least one of the numbers must be zero. So, either $a = 0$ or $1 - b = 0$.

Case 1: $a = 0$.

If $a=0$, the equation $a(1-b)=0$ is satisfied for any value of $b$. Since $a$ must be a whole number, $a=0$ is a valid value for $a$. We are given that $b$ is a whole number greater than 1 ($b \in \{2, 3, 4, ...\}$). If we choose $a=0$ and any $b > 1$ (for example, $b=3$), the conditions are met ($a=0$ is a whole number, $b=3$ is a whole number greater than 1).

Let's check the original division for this case:

Here, the quotient is $0$. The former number is $a=0$. The quotient ($0$) is equal to the former number ($0$). So, the condition $\frac{a}{b} = a$ holds when $a=0$ and $b>1$.

Case 2: $1 - b = 0$, which means $b = 1$.

For the equation $a(1-b)=0$ to hold, $b$ could be 1. However, the problem statement requires that the second whole number $b$ must be strictly greater than 1 ($b > 1$). Therefore, $b=1$ is not allowed under the conditions of the problem.

From the analysis, we found that the quotient is equal to the former number when $a=0$ and $b$ is any whole number greater than 1.

For example, if $a=0$ and $b=5$, $a$ is a whole number, $b$ is a whole number greater than 1 ($5 > 1$). The quotient $\frac{a}{b} = \frac{0}{5} = 0$. The former number is $a=0$. The quotient ($0$) is equal to the former number ($0$).

Since we found a case where the quotient *is* equal to the former number (when the former number is $0$), the statement that the quotient "never" gives the quotient equal to the former is false.

Answer:

The statement is False.

Given:

A number and its multiples.

To Check:

The truth value of the statement: "Every multiple of a number is greater than or equal to the number".

Solution:

Let the number be $n$. We will consider $n$ as a whole number, as per the context of preceding questions.

So, $n \in \{0, 1, 2, 3, ...\}$.

A multiple of $n$ is obtained by multiplying $n$ by a whole number $k$, where $k \in \{0, 1, 2, 3, ...\}$.

The multiples of $n$ are of the form $n \times k$.

The statement claims that for any whole number $n$ and any whole number $k$, the multiple $n \times k$ is greater than or equal to $n$.

Let's examine this statement by considering different values for $n$ and $k$.

Consider a non-zero whole number, for example, let $n = 5$.

$5$ is a whole number.

Its multiples are $5 \times k$ for $k \in \{0, 1, 2, 3, ...\}$.

For $k=0$, the multiple is $5 \times 0 = 0$.

The statement claims this multiple ($0$) should be greater than or equal to the number ($5$).

So, we check if $0 \ge 5$.

This is clearly false, as $0$ is less than $5$.

Therefore, for the number $5$, the multiple $0$ is not greater than or equal to $5$.

This single counterexample is sufficient to prove that the statement "Every multiple of a number is greater than or equal to the number" is false.

Let's consider the case where $n=0$. The multiples of $0$ are $0 \times k$. For any whole number $k$, $0 \times k = 0$. The statement claims that every multiple ($0$) is greater than or equal to the number ($0$), i.e., $0 \ge 0$. This is true. So the statement holds for $n=0$.

However, the statement says "Every multiple of a number is greater than or equal to the number". This must hold for any number (within the defined set, which we assume to be whole numbers). Since it does not hold for non-zero whole numbers like $5$ (where the multiple $0$ is not $\ge 5$), the overall statement is false.

Answer:

The statement is False.

Given:

A given number.

To Check:

The truth value of the statement: "The number of multiples of a given number is finite".

Solution:

Let the given number be $n$. Assuming the context of whole numbers, $n$ is a whole number, so $n \in \{0, 1, 2, 3, ...\}$.

A multiple of $n$ is obtained by multiplying $n$ by any whole number $k$. That is, a multiple is of the form $n \times k$, where $k \in \{0, 1, 2, 3, ...\}$.

We need to determine if the set of such multiples $\{nk \mid k \in \{0, 1, 2, 3, ...\}\}$ is a finite set.

Let's consider two cases for the given number $n$:

Case 1: The given number $n$ is 0.

The multiples of $0$ are $0 \times k$ for $k \in \{0, 1, 2, 3, ...\}$.

$0 \times 0 = 0$

$0 \times 1 = 0$

$0 \times 2 = 0$

... and so on.

The set of multiples of $0$ is $\{0\}$, which contains only one element. This is a finite set.

Case 2: The given number $n$ is a non-zero whole number.

This means $n \in \{1, 2, 3, ...\}$.

The multiples of $n$ are $n \times k$ for $k \in \{0, 1, 2, 3, ...\}$.

For $k=0$, the multiple is $n \times 0 = 0$.

For $k=1$, the multiple is $n \times 1 = n$.

For $k=2$, the multiple is $n \times 2 = 2n$.

For $k=3$, the multiple is $n \times 3 = 3n$.

... and so on.

The set of multiples is $\{0, n, 2n, 3n, 4n, ...\}$.

Since $n$ is a non-zero whole number ($n \ge 1$), each time we multiply $n$ by a different whole number $k$ ($k=0, 1, 2, 3, ...$), we get a distinct multiple. For example, if $n=5$, the multiples are $0, 5, 10, 15, 20, ...$. These are all distinct values.

The set of whole numbers $\{0, 1, 2, 3, ...\}$ is an infinite set.

Since we are multiplying $n$ (where $n \neq 0$) by every number in this infinite set, and each multiplication produces a unique value, the resulting set of multiples $\{0, n, 2n, 3n, ...\}$ is also infinite.

The statement claims that "The number of multiples of a given number is finite". This implies it should hold true for any whole number. However, we found that for any non-zero whole number, the number of multiples is infinite.

Therefore, the statement is false because it does not hold for all (non-zero) whole numbers.

Answer:

The statement is False.

Given:

A number.

To Check:

The truth value of the statement: "Every number is a multiple of itself".

Solution:

We assume the context of the question refers to whole numbers, given the preceding questions.

Let the number be $n$, where $n$ is a whole number ($n \in \{0, 1, 2, 3, ...\}$).

By definition, a number $m$ is a multiple of another number $n$ if $m = n \times k$ for some whole number $k$ ($k \in \{0, 1, 2, 3, ...\}$).

The statement says that every number $n$ is a multiple of itself. This means we need to check if for any whole number $n$, there exists a whole number $k$ such that the following equation holds:

We consider two cases for the value of $n$:

Case 1: The number $n = 0$

If $n=0$, the equation becomes $0 = 0 \times k$.

We need to determine if there is at least one whole number $k$ that satisfies this equation.

For any whole number $k$ (e.g., $k=0, 1, 2, 3, ...$), the product $0 \times k$ is always $0$.

So, $0 = 0 \times k$ is true for any whole number $k$. This confirms that $0$ is a multiple of $0$.

Case 2: The number $n$ is a non-zero whole number

If $n \neq 0$, where $n \in \{1, 2, 3, ...\}$, the equation is $n = n \times k$.

We need to find if there exists a whole number $k$ satisfying this equation.

Since $n$ is a non-zero number, we can divide both sides of the equation by $n$:

The value of $k$ that satisfies the equation is $1$. We need to check if $1$ is a whole number. Yes, $1$ is a whole number.

So, for any non-zero whole number $n$, we can choose $k=1$ to satisfy the condition $n = n \times k$. This means any non-zero whole number is a multiple of itself.

Since the statement holds true for $n=0$ and also for all non-zero whole numbers, it holds for every whole number.

Answer:

The statement is True.

Given:

Sum of two consecutive odd numbers.

To Check:

The truth value of the statement: "Sum of two consecutive odd numbers is always divisible by 4".

Solution:

Let the first odd number be represented by $2n + 1$, where $n$ is a whole number ($n \in \{0, 1, 2, 3, ...\}$). This formula generates all odd numbers: $2(0)+1=1$, $2(1)+1=3$, $2(2)+1=5$, and so on.

The next consecutive odd number is obtained by adding 2 to the first odd number. So, the second consecutive odd number is $(2n + 1) + 2 = 2n + 3$.

We need to find the sum of these two consecutive odd numbers:

Sum $= (2n + 1) + (2n + 3)$

Combine like terms:

Now, we can factor out a common factor of 4 from the sum:

The sum is expressed as $4$ multiplied by the term $(n + 1)$.

Since $n$ is a whole number ($n \in \{0, 1, 2, 3, ...\}$), the term $(n + 1)$ is also a whole number ($n+1 \in \{1, 2, 3, 4, ...\}$).

A number is divisible by 4 if it can be written in the form $4 \times m$, where $m$ is a whole number.

In our case, the sum is $4 \times (n+1)$, and $(n+1)$ is a whole number.

This shows that the sum of any two consecutive odd numbers is always a multiple of 4.

Being a multiple of 4 means being divisible by 4.

Let's test with a few examples:

• Consecutive odd numbers: 1 and 3. Sum = $1 + 3 = 4$. $4$ is divisible by $4$ ($4 = 4 \times 1$). (Here $n=0$, sum $= 4(0+1)=4$)
• Consecutive odd numbers: 3 and 5. Sum = $3 + 5 = 8$. $8$ is divisible by $4$ ($8 = 4 \times 2$). (Here $n=1$, sum $= 4(1+1)=8$)
• Consecutive odd numbers: 11 and 13. Sum = $11 + 13 = 24$. $24$ is divisible by $4$ ($24 = 4 \times 6$). (Here $n=5$, sum $= 4(5+1)=24$)

In every case, the sum is a multiple of 4, and thus divisible by 4.

The algebraic representation $4(n+1)$ confirms that the sum is always a multiple of 4 for any whole number $n$, which generates all pairs of consecutive odd numbers.

Answer:

The statement is True.

Given:

A number divides three numbers exactly.

To Check:

The truth value of the statement: "it must divide their sum exactly".

Solution:

Let the number that divides be $d$.

Let the three numbers be $a$, $b$, and $c$.

The statement "a number divides three numbers exactly" means that $d$ is a divisor of $a$, $d$ is a divisor of $b$, and $d$ is a divisor of $c$.

This implies that $a$, $b$, and $c$ can be written as multiples of $d$ with zero remainders.

So, there exist some whole numbers $k_1$, $k_2$, and $k_3$ such that:

We need to check if $d$ divides the sum of these three numbers exactly. The sum is $a + b + c$.

Substitute the expressions for $a$, $b$, and $c$ into the sum:

Sum $= a + b + c = (d \times k_1) + (d \times k_2) + (d \times k_3)$

Now, we can factor out the common factor $d$ from the sum:

Let $K = k_1 + k_2 + k_3$. Since $k_1$, $k_2$, and $k_3$ are whole numbers, their sum $K$ is also a whole number.

So, the sum can be written as:

This means that the sum $a+b+c$ is a multiple of $d$.

By definition, if a number can be written as a multiple of $d$ with a whole number multiplier ($K$ in this case), then $d$ divides that number exactly.

Thus, $d$ divides the sum $(a+b+c)$ exactly.

This property holds true for any number $d$ that exactly divides $a$, $b$, and $c$. It is a fundamental property of divisibility.

Let's consider an example:

Let the number be $d=3$. Let the three numbers be $a=6$, $b=9$, and $c=15$.

Check if $d$ divides $a, b, c$ exactly:

$6 \div 3 = 2$ (exactly, $6 = 3 \times 2$)

$9 \div 3 = 3$ (exactly, $9 = 3 \times 3$)

$15 \div 3 = 5$ (exactly, $15 = 3 \times 5$)

The sum is $a + b + c = 6 + 9 + 15 = 30$.

Check if $d$ divides the sum exactly:

$30 \div 3 = 10$ (exactly, $30 = 3 \times 10$)

The sum is indeed divisible by $d=3$.

This demonstrates the property. The algebraic proof confirms it holds generally.

Answer:

The statement is True.

Given:

A number exactly divides the sum of three numbers.

To Check:

The truth value of the statement: "it must exactly divide the numbers separately".

Solution:

Let the number that divides be $d$.

Let the three numbers be $a$, $b$, and $c$.

The given condition is that $d$ exactly divides the sum $(a+b+c)$.

This means that $a+b+c = d \times K$ for some whole number $K$.

The statement claims that this implies $d$ must exactly divide $a$, $d$ must exactly divide $b$, and $d$ must exactly divide $c$.

To verify this statement, we can try to find a counterexample. A counterexample is a specific instance where the given condition is true, but the conclusion is false.

We need to find a number $d$ and three numbers $a, b, c$ such that:

1. $d$ exactly divides $a+b+c$.

2. $d$ does not exactly divide at least one of $a$, $b$, or $c$.

Let's choose a simple example.

Let the number $d = 5$.

Let the three numbers be $a = 1$, $b = 2$, and $c = 2$.

Check the condition: Find the sum of the three numbers.

Now, check if $d=5$ exactly divides the sum $5$.

$5 \div 5 = 1$. The quotient is a whole number (1) and the remainder is 0. So, $5$ exactly divides $5$.

The given condition holds true for this example.

Now, check the conclusion: Does $d=5$ exactly divide the numbers $a$, $b$, and $c$ separately?

Check if $5$ exactly divides $a=1$:

$1 \div 5$. The quotient is $0$ and the remainder is $1$. $5$ does not exactly divide $1$.

Since we have found that $d$ does not exactly divide $a$, the conclusion of the statement is false for this example.

Another example:

Let $d=3$. Let $a=1$, $b=1$, $c=1$.

Sum $= 1+1+1 = 3$. $d=3$ divides the sum $3$ exactly ($3 \div 3 = 1$).

Does $d=3$ exactly divide $a=1$, $b=1$, $c=1$ separately? No, $3$ does not divide $1$ exactly.

Since we can find instances where a number divides the sum of three numbers exactly, but does not divide the numbers separately, the statement is not universally true.

Answer:

The statement is False.

Given:

A number is divisible both by 2 and 3.

To Check:

The truth value of the statement: "then it is divisible by 12".

Solution:

Let the number be $N$.

The given condition is that $N$ is divisible by 2 and $N$ is divisible by 3.

If a number is divisible by 2, it means $N$ is a multiple of 2. So, $N = 2 \times k_1$ for some integer $k_1$.

If a number is divisible by 3, it means $N$ is a multiple of 3. So, $N = 3 \times k_2$ for some integer $k_2$.

A number is divisible by 2 and 3 if and only if it is divisible by the least common multiple (LCM) of 2 and 3.

Let's find the LCM of 2 and 3.

Prime factorization of 2: $2 = 2^1$

Prime factorization of 3: $3 = 3^1$

The LCM is found by taking the highest power of all prime factors involved:

So, a number divisible by both 2 and 3 is divisible by 6.

This means $N$ must be a multiple of 6. So, $N = 6 \times m$ for some integer $m$.

The statement claims that if $N$ is divisible by 6, then it must be divisible by 12.

Let's check if every multiple of 6 is also a multiple of 12.

Multiples of 6 are: $0, 6, 12, 18, 24, 30, 36, ...$

Multiples of 12 are: $0, 12, 24, 36, ...$

We can see that the multiples of 6 include numbers that are not multiples of 12, such as 6, 18, 30, etc.

For example, consider the number 6.

6 is divisible by 2 ($6 \div 2 = 3$).

6 is divisible by 3 ($6 \div 3 = 2$).

So, 6 satisfies the given condition: "A number is divisible both by 2 and 3".

Now, check the conclusion: Is 6 divisible by 12?

$6 \div 12 = 0$ with a remainder of $6$. 6 is not divisible by 12 exactly.

Since we found a number (6) that is divisible by both 2 and 3, but is not divisible by 12, the statement is false.

A number is divisible by $a$ and $b$ if and only if it is divisible by LCM$(a,b)$. If $\text{LCM}(a,b) = ab$, then the number is divisible by $ab$. If $\text{LCM}(a,b) \ne ab$, then the number is divisible by $\text{LCM}(a,b)$, not necessarily by $ab$. In this case, $\text{LCM}(2,3) = 6$, and $2 \times 3 = 6$. So a number divisible by 2 and 3 is divisible by 6.

For a number to be divisible by both 2 and 3, it must be a multiple of LCM(2, 3) = 6.

For a number to be divisible by 12, it must be a multiple of 12.

The statement is essentially saying: "If a number is a multiple of 6, then it is a multiple of 12". This is false, as shown by the counterexample 6.

Answer:

The statement is False.

Given:

A number with three or more digits. The number formed by its last two digits is divisible by 6.

To Check:

The truth value of the statement: "the number is divisible by 6".

Solution:

Let the number with three or more digits be $N$.

We can write $N$ in expanded form based on its digits. For a number with $d$ digits ($d \ge 3$), let the digits be $a_{d-1} a_{d-2} ... a_2 a_1 a_0$, where $a_0$ is the units digit and $a_1$ is the tens digit.

The number can be written as:

We can separate the last two digits ($a_1 a_0$) from the rest of the number:

Let $M$ be the number formed by the digits from the hundreds place onwards: $M = a_{d-1} \times 10^{d-3} + ... + a_2$. Note that this part is multiplied by 100.

So, we can write $N$ as:

Let $L$ be the number formed by the last two digits: $L = 10 a_1 + a_0$.

So, $N = 100M + L$.

The given condition is that the number formed by the last two digits ($L$) is divisible by 6.

This means $L$ is a multiple of 6, so $L = 6k$ for some integer $k$. Since $L$ is formed by two digits, $0 \le L \le 99$. If $L$ is divisible by 6, $L$ can be $00, 06, 12, 18, ..., 96$. These are multiples of 6.

A number is divisible by 6 if and only if it is divisible by both 2 and 3.

Since $L$ is divisible by 6, $L$ is divisible by 2 and $L$ is divisible by 3.

Consider the number $N = 100M + L$.

For $N$ to be divisible by 6, $N$ must be divisible by 2 and 3.

Divisibility by 2: A number is divisible by 2 if its last digit is even. The last digit of $N$ is $a_0$, which is the last digit of $L$. Since $L$ is divisible by 6, $L$ must be even (as all multiples of 6 are even). Therefore, the last digit $a_0$ must be even. This means $N$ is divisible by 2.

Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of the digits of $N$ is $a_{d-1} + ... + a_2 + a_1 + a_0$. The sum of the digits of $L$ is $a_1 + a_0$. Since $L$ is divisible by 6, $L$ is divisible by 3. However, this does not mean the sum of the digits of $L$ is divisible by 3. For example, if $L=12$, the sum of digits is $1+2=3$, which is divisible by 3. If $L=24$, the sum of digits is $2+4=6$, divisible by 3. If $L=48$, the sum of digits is $4+8=12$, divisible by 3.

Let's consider the property that a number is divisible by 3 if and only if the sum of its digits is divisible by 3. This is a valid divisibility rule for 3.

The sum of the digits of $N$ is $S_N = (a_{d-1} + ... + a_2) + (a_1 + a_0)$.

The sum of the digits of $L$ is $S_L = a_1 + a_0$.

We know that $L = 10a_1 + a_0$ is divisible by 6, which means $10a_1 + a_0$ is divisible by 3.

Using modular arithmetic, $10 \equiv 1 \pmod{3}$, so $10a_1 + a_0 \equiv 1a_1 + a_0 \equiv a_1 + a_0 \pmod{3}$.

Since $10a_1 + a_0$ is divisible by 3, we have $10a_1 + a_0 \equiv 0 \pmod{3}$.

Therefore, $a_1 + a_0 \equiv 0 \pmod{3}$. This means the sum of the last two digits ($a_1 + a_0$) is divisible by 3. This is always true if $L$ is divisible by 3 (which it is, since $L$ is divisible by 6).

However, the divisibility of $N$ by 3 depends on the sum of all its digits $S_N = (a_{d-1} + ... + a_2) + (a_1 + a_0)$.

We know $a_1 + a_0$ is divisible by 3. But we don't know if $(a_{d-1} + ... + a_2)$ is divisible by 3.

Let's try a counterexample.

We need a number $N \ge 100$ such that the number formed by the last two digits is divisible by 6, but $N$ itself is not divisible by 6.

Let the last two digits form the number $L = 12$. $12$ is divisible by 6 ($12 = 6 \times 2$).

We need to choose the preceding digits such that the resulting number is not divisible by 6.

Consider $N = 112$. The last two digits form 12, which is divisible by 6.

Check if 112 is divisible by 6. For divisibility by 6, it must be divisible by both 2 and 3.

112 is divisible by 2 because its last digit is 2 (even).

Check divisibility by 3: Sum of digits $= 1 + 1 + 2 = 4$. 4 is not divisible by 3.

Since 112 is not divisible by 3, it is not divisible by 6.

In this example, the number formed by the last two digits (12) is divisible by 6, but the number (112) is not divisible by 6.

This contradicts the statement.

The divisibility rule for 6 requires divisibility by both 2 (checked by the last digit) and 3 (checked by the sum of all digits). The divisibility of the last two digits by 6 only guarantees divisibility by 2 (of the whole number) and divisibility by 3 of the sum of the last two digits. It does not guarantee divisibility by 3 of the sum of all digits.

Answer:

The statement is False.

Given:

A number with 4 or more digits. The number formed by its last three digits is divisible by 8.

To Check:

The truth value of the statement: "the number is divisible by 8".

Solution:

Let the number with 4 or more digits be $N$.

We can write $N$ in expanded form based on its digits. For a number with $d$ digits ($d \ge 4$), let the digits be $a_{d-1} a_{d-2} ... a_3 a_2 a_1 a_0$, where $a_0$ is the units digit, $a_1$ is the tens digit, and $a_2$ is the hundreds digit.

The number can be written as:

We can separate the last three digits ($a_2 a_1 a_0$) from the rest of the number:

Let $M$ be the number formed by the digits from the thousands place onwards: $M = a_{d-1} \times 10^{d-4} + ... + a_3$. Note that this part is multiplied by 1000.

So, we can write $N$ as:

Let $L$ be the number formed by the last three digits: $L = 100 a_2 + 10 a_1 + a_0$.

So, $N = 1000M + L$.

The given condition is that the number formed by the last three digits ($L$) is divisible by 8.

This means $L = 8k$ for some integer $k$.

We want to check if $N$ is divisible by 8.

We have $N = 1000M + L$.

Consider the term $1000M$. We know that $1000$ is divisible by $8$ ($1000 = 8 \times 125$).

So, $1000M = (8 \times 125) \times M = 8 \times (125M)$.

Since $M$ is an integer (formed by digits), $125M$ is also an integer. This means $1000M$ is always a multiple of 8, and thus divisible by 8.

Now, consider the sum $N = 1000M + L$.

We know that $1000M$ is divisible by 8.

We are given that $L$ is divisible by 8.

If two numbers are divisible by a number $d$, their sum is also divisible by $d$.

Since $1000M$ is divisible by 8 and $L$ is divisible by 8, their sum $1000M + L$ must be divisible by 8.

Since $L$ is divisible by 8, $\frac{L}{8}$ is an integer. Since $M$ is an integer, $125M$ is an integer. The sum of two integers ($125M + \frac{L}{8}$) is an integer.

This shows that $N$ divided by 8 results in an integer, meaning $N$ is divisible by 8.

This property is the actual divisibility rule for 8 for numbers with 3 or more digits.

Let's consider an example:

Let $N = 5120$. This number has 4 digits.

The number formed by the last three digits is 120.

Check if 120 is divisible by 8: $120 \div 8 = 15$. Yes, 120 is divisible by 8.

According to the statement, 5120 should be divisible by 8.

Check if 5120 is divisible by 8: $5120 \div 8 = 640$. Yes, 5120 is divisible by 8.

Another example:

Let $N = 7048$. This number has 4 digits.

The number formed by the last three digits is 048, which is 48.

Check if 48 is divisible by 8: $48 \div 8 = 6$. Yes, 48 is divisible by 8.

According to the statement, 7048 should be divisible by 8.

Check if 7048 is divisible by 8: $7048 \div 8 = 881$. Yes, 7048 is divisible by 8.

The statement is consistent with the divisibility rule for 8.

Answer:

The statement is True.

Given:

The sum of the digits of a number is divisible by 3.

To Check:

The truth value of the statement: "then the number itself is divisible by 9".

Solution:

Let the number be $N$. Let the digits of the number be $d_k d_{k-1} ... d_1 d_0$.

The number $N$ can be written as:

The sum of the digits is $S = d_k + d_{k-1} + ... + d_1 + d_0$.

The given condition is that the sum of the digits $S$ is divisible by 3.

This is the divisibility rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

So, the given condition means that the number $N$ is divisible by 3.

The statement claims that if $N$ is divisible by 3, then $N$ must be divisible by 9.

Let's check if every number that is divisible by 3 is also divisible by 9.

Consider the multiples of 3:

$3, 6, 9, 12, 15, 18, 21, 24, 27, ...$

Consider the multiples of 9:

$9, 18, 27, 36, ...$

We can see from the list of multiples of 3 that some numbers are divisible by 3 but not by 9. For example:

• The number 3: Sum of digits is 3. 3 is divisible by 3. Is 3 divisible by 9? No.
• The number 6: Sum of digits is 6. 6 is divisible by 3. Is 6 divisible by 9? No.
• The number 12: Sum of digits is $1+2=3$. 3 is divisible by 3. Is 12 divisible by 9? No.

Let's verify the condition and conclusion for one of these numbers as a counterexample.

Consider the number $N = 12$.

The sum of the digits of 12 is $1 + 2 = 3$.

Is the sum of digits (3) divisible by 3? Yes, $3 \div 3 = 1$. The given condition is met.

Now, check the conclusion: Is the number itself (12) divisible by 9?

$12 \div 9 = 1$ with a remainder of $3$. 12 is not divisible by 9 exactly.

Since we found a number (12) for which the sum of the digits is divisible by 3, but the number itself is not divisible by 9, the statement is false.

The divisibility rule for 9 is: A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

If the sum of digits is divisible by 3, it only guarantees divisibility by 3, not necessarily by 9 (unless the sum of digits is also a multiple of 9, which is a stronger condition).

Answer:

The statement is False.

Given:

Numbers which are divisible by 4.

To Check:

The truth value of the statement: "All numbers which are divisible by 4 may not be divisible by 8".

Solution:

The statement says that it is possible for a number to be divisible by 4, but not divisible by 8. This means the set of numbers divisible by 4 is not completely contained within the set of numbers divisible by 8.

Let's list some numbers divisible by 4:

These are multiples of 4: $0, 4, 8, 12, 16, 20, 24, 28, 32, ...$

Let's list some numbers divisible by 8:

These are multiples of 8: $0, 8, 16, 24, 32, ...$

The statement "All numbers which are divisible by 4 may not be divisible by 8" is equivalent to stating that there exists at least one number which is divisible by 4, but is not divisible by 8.

Let's examine the list of multiples of 4:

• Is 0 divisible by 4? Yes ($0 = 4 \times 0$). Is 0 divisible by 8? Yes ($0 = 8 \times 0$).
• Is 4 divisible by 4? Yes ($4 = 4 \times 1$). Is 4 divisible by 8? No ($4 \div 8 = 0$ with remainder 4).

We have found a number, 4, which is divisible by 4, but is not divisible by 8.

This single example is enough to support the statement.

Consider other numbers from the list of multiples of 4:

• 12 is divisible by 4 ($12 = 4 \times 3$). Is 12 divisible by 8? No ($12 \div 8 = 1$ with remainder 4).
• 20 is divisible by 4 ($20 = 4 \times 5$). Is 20 divisible by 8? No ($20 \div 8 = 2$ with remainder 4).
• 28 is divisible by 4 ($28 = 4 \times 7$). Is 28 divisible by 8? No ($28 \div 8 = 3$ with remainder 4).

In general, a number divisible by 4 can be written as $4k$ for some integer $k$.

For this number $4k$ to be divisible by 8, it must be a multiple of 8, i.e., $4k = 8m$ for some integer $m$.

Dividing by 4, we get $k = 2m$.

This means that $k$ must be an even integer. If $k$ is an odd integer, then $4k$ will be divisible by 4 but not by 8.

For example, if $k=1$, the number is $4 \times 1 = 4$. $k=1$ is odd, and 4 is divisible by 4 but not 8.

If $k=3$, the number is $4 \times 3 = 12$. $k=3$ is odd, and 12 is divisible by 4 but not 8.

If $k=5$, the number is $4 \times 5 = 20$. $k=5$ is odd, and 20 is divisible by 4 but not 8.

The statement implies that the set of numbers divisible by 4 is larger than the set of numbers divisible by 8, in the sense that there are elements in the former set not present in the latter. This is true.

Answer:

The statement is True.

Given:

Two or more numbers.

To Check:

The truth value of the statement: "The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple".

Solution:

Let's consider two positive integers $a$ and $b$.

The Highest Common Factor (HCF), also known as the Greatest Common Divisor (GCD), is the largest positive integer that divides both $a$ and $b$ without leaving a remainder.

By definition, the HCF of $a$ and $b$, denoted as HCF$(a, b)$, must divide both $a$ and $b$. This implies that HCF$(a, b) \le a$ and HCF$(a, b) \le b$. Consequently, HCF$(a, b)$ is less than or equal to the smallest of the numbers ($a$ or $b$).

The Lowest Common Multiple (LCM) is the smallest positive integer that is a multiple of both $a$ and $b$.

By definition, the LCM of $a$ and $b$, denoted as LCM$(a, b)$, must be a multiple of both $a$ and $b$. This implies that LCM$(a, b) \ge a$ and LCM$(a, b) \ge b$. Consequently, LCM$(a, b)$ is greater than or equal to the largest of the numbers ($a$ or $b$).

Since for positive numbers, $\min(a, b) \le \max(a, b)$, we can generally say that:

This relationship shows that for positive numbers, the HCF is always less than or equal to the LCM. The only case where HCF = LCM is when the numbers are equal (and non-zero).

The statement claims that HCF is greater than LCM. This contradicts the fundamental relationship HCF $\le$ LCM.

Let's consider examples:

Example 1: Numbers are 6 and 9.

Factors of 6: 1, 2, 3, 6

Factors of 9: 1, 3, 9

Common factors: 1, 3. HCF(6, 9) = 3.

Multiples of 6: 0, 6, 12, 18, 24, ...

Multiples of 9: 0, 9, 18, 27, ...

Common multiples (excluding 0): 18, 36, ... Lowest positive common multiple is 18. LCM(6, 9) = 18.

Here, HCF(6, 9) = 3 and LCM(6, 9) = 18. $3 \le 18$. The statement $3 > 18$ is false.

Example 2: Numbers are 4 and 8.

Factors of 4: 1, 2, 4

Factors of 8: 1, 2, 4, 8

Common factors: 1, 2, 4. HCF(4, 8) = 4.

Multiples of 4: 0, 4, 8, 12, 16, ...

Multiples of 8: 0, 8, 16, 24, ...

Common multiples (excluding 0): 8, 16, 24, ... Lowest positive common multiple is 8. LCM(4, 8) = 8.

Here, HCF(4, 8) = 4 and LCM(4, 8) = 8. $4 \le 8$. The statement $4 > 8$ is false.

Example 3: Numbers are 5 and 5.

Factors of 5: 1, 5. HCF(5, 5) = 5.

Multiples of 5: 0, 5, 10, 15, ... LCM(5, 5) = 5.

Here, HCF(5, 5) = 5 and LCM(5, 5) = 5. $5 = 5$. The statement $5 > 5$ is false.

In all cases with positive numbers, HCF is less than or equal to LCM. The statement claims the opposite.

The only exception might involve the number 0, but HCF and LCM are typically defined for positive integers or at least non-negative integers where special rules apply for 0 (e.g., HCF(a,0)=a, LCM(a,0)=0). Assuming the question is about positive integers, the statement is definitively false.

Answer:

The statement is False.

Given:

Two or more numbers, their HCF, and their LCM.

To Check:

The truth value of the statement: "LCM of two or more numbers is divisible by their HCF".

Solution:

Let's consider two positive integers $a$ and $b$.

Let HCF$(a, b) = h$ and LCM$(a, b) = l$.

By definition, $h$ is the largest positive integer that divides both $a$ and $b$.

By definition, $l$ is the smallest positive integer that is a multiple of both $a$ and $b$.

A fundamental property relating HCF and LCM of two positive integers $a$ and $b$ is:

Using our notation, this is $ab = hl$.

From this relationship, we can express $l$ in terms of $a$, $b$, and $h$ (assuming $h \ne 0$, which is true for positive $a, b$):

We know that $h$ divides $a$ and $h$ divides $b$.

Let $a = h \times x$ and $b = h \times y$, where $x$ and $y$ are integers. Also, since $h$ is the HCF, $x$ and $y$ must be coprime (HCF$(x, y) = 1$).

Now substitute these into the expression for $l$:

So, LCM$(a, b) = h \times (xy)$.

Since $x$ and $y$ are integers (and positive if $a, b, h$ are positive), $xy$ is an integer.

The equation $l = h \times (xy)$ shows that $l$ is a multiple of $h$.

If $l$ is a multiple of $h$, it means $l$ is divisible by $h$ exactly.

This proves the statement for two numbers.

The property also extends to more than two numbers. While the direct product formula $a \times b \times c = \text{HCF}(a, b, c) \times \text{LCM}(a, b, c)$ is not generally true for three numbers, the relationship that HCF divides LCM still holds.

Consider numbers $a_1, a_2, ..., a_n$. Let $h = \text{HCF}(a_1, ..., a_n)$ and $l = \text{LCM}(a_1, ..., a_n)$.

By definition of HCF, $h$ divides every $a_i$. So, $a_i = h k_i$ for some integers $k_i$.

By definition of LCM, $l$ is a multiple of every $a_i$. So, $l = m_i a_i$ for some integers $m_i$.

To show that $l$ is divisible by $h$, we need to show that $\frac{l}{h}$ is an integer.

Every prime factor of $h$ must be a prime factor of every $a_i$. The exponent of any prime factor in $h$ is the minimum of its exponents in the prime factorization of each $a_i$.

Every prime factor of $l$ is a prime factor of at least one $a_i$. The exponent of any prime factor in $l$ is the maximum of its exponents in the prime factorization of each $a_i$.

Let $p$ be a prime factor. Let $p^{\alpha_i}$ be the highest power of $p$ that divides $a_i$.

The highest power of $p$ that divides $h$ is $p^{\min(\alpha_1, ..., \alpha_n)}$.

The highest power of $p$ that divides $l$ is $p^{\max(\alpha_1, ..., \alpha_n)}$.

For $l$ to be divisible by $h$, the exponent of every prime factor $p$ in $l$ must be greater than or equal to its exponent in $h$.

This means we need to check if $\max(\alpha_1, ..., \alpha_n) \ge \min(\alpha_1, ..., \alpha_n)$.

This inequality is always true, as the maximum of a set of numbers is always greater than or equal to the minimum of that set.

Since this holds for every prime factor, $l$ is divisible by $h$.

Let's verify with an example with three numbers: 12, 18, 30.

Prime factorizations:

$12 = 2^2 \times 3^1 \times 5^0$

$18 = 2^1 \times 3^2 \times 5^0$

$30 = 2^1 \times 3^1 \times 5^1$

HCF(12, 18, 30): Take the minimum exponent for each prime factor.

For 2: $\min(2, 1, 1) = 1$. So, $2^1$ is in HCF.

For 3: $\min(1, 2, 1) = 1$. So, $3^1$ is in HCF.

For 5: $\min(0, 0, 1) = 0$. So, $5^0$ is in HCF.

HCF(12, 18, 30) $= 2^1 \times 3^1 \times 5^0 = 2 \times 3 \times 1 = 6$.

LCM(12, 18, 30): Take the maximum exponent for each prime factor.

For 2: $\max(2, 1, 1) = 2$. So, $2^2$ is in LCM.

For 3: $\max(1, 2, 1) = 2$. So, $3^2$ is in LCM.

For 5: $\max(0, 0, 1) = 1$. So, $5^1$ is in LCM.

LCM(12, 18, 30) $= 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180$.

Check if LCM (180) is divisible by HCF (6):

$180 \div 6 = 30$. Yes, 180 is divisible by 6.

The property holds for more than two numbers as well.

Answer:

The statement is True.

Given:

Proposed values for the LCM and HCF of two numbers: LCM = 28 and HCF = 8.

To Check:

The truth value of the statement, which implicitly asks if it is possible for two numbers to have LCM = 28 and HCF = 8.

Solution:

We know that for any two positive integers $a$ and $b$, their Lowest Common Multiple (LCM) and Highest Common Factor (HCF) are related by the property that the HCF must divide the LCM exactly.

This relationship can be stated as:

Or, equivalently,

In the given statement, we are provided with the proposed values: LCM = 28 and HCF = 8.

We need to check if the HCF (8) divides the LCM (28) exactly.

Divide 28 by 8:

Let's perform the division:

Since the remainder is 4 (not 0), 28 is not divisible by 8.

The property that LCM must be divisible by HCF is a fundamental requirement for any pair of numbers. Since the given proposed values (LCM = 28, HCF = 8) violate this property (28 is not divisible by 8), it is impossible for two numbers to have this combination of HCF and LCM.

Therefore, the statement describing this situation is false because such a pair of numbers cannot exist.

The reason behind the property HCF divides LCM stems from the prime factorization of numbers. Let the prime factorization of two numbers $a$ and $b$ be:

where $p_i$ are prime numbers and $\alpha_i, \beta_i \ge 0$ are integers.

The HCF is given by:

The LCM is given by:

For HCF to divide LCM, the exponent of each prime $p_i$ in HCF must be less than or equal to the exponent of $p_i$ in LCM.

This means we need $\min(\alpha_i, \beta_i) \le \max(\alpha_i, \beta_i)$ for all $i$. This inequality is always true.

In our case:

HCF = 8 = $2^3$

LCM = 28 = $2^2 \times 7^1$

Comparing the prime factorizations:

For the prime factor 2: The exponent in HCF is 3, the exponent in LCM is 2. We need $3 \le 2$ for HCF to divide LCM, which is false.

For the prime factor 7: The exponent in HCF is 0 (since 8 has no factor of 7), the exponent in LCM is 1. We need $0 \le 1$, which is true.

Since the condition is not met for the prime factor 2, HCF (8) does not divide LCM (28).

Answer:

The statement is False.

Given:

Two or more numbers and their LCM.

To Check:

The truth value of the statement: "LCM of two or more numbers may be one of the numbers".

Solution:

Let's consider a set of two or more positive integers: $\{a_1, a_2, ..., a_n\}$, where $n \ge 2$.

The Lowest Common Multiple (LCM) of these numbers, denoted as LCM$(a_1, ..., a_n)$, is the smallest positive integer that is a multiple of each number in the set.

By definition, for any number $a_i$ in the set, LCM$(a_1, ..., a_n)$ must be a multiple of $a_i$.

We need to check if it is possible for the LCM to be equal to one of the numbers in the set.

Let's consider some examples:

Example 1: Consider the numbers 4 and 8.

Multiples of 4: 0, 4, 8, 12, 16, ...

Multiples of 8: 0, 8, 16, 24, ...

The common multiples are $0, 8, 16, 24, ...$. The lowest positive common multiple is 8.

LCM(4, 8) = 8.

In this case, the LCM (8) is equal to one of the numbers (8) in the set {4, 8}.

Example 2: Consider the numbers 6 and 12.

Multiples of 6: 0, 6, 12, 18, ...

Multiples of 12: 0, 12, 24, ...

The common multiples are $0, 12, 24, ...$. The lowest positive common multiple is 12.

LCM(6, 12) = 12.

In this case, the LCM (12) is equal to one of the numbers (12) in the set {6, 12}.

Example 3: Consider the numbers 3, 6, and 12.

Multiples of 3: 0, 3, 6, 9, 12, 15, ...

Multiples of 6: 0, 6, 12, 18, ...

Multiples of 12: 0, 12, 24, ...

The common multiples are $0, 12, 24, ...$. The lowest positive common multiple is 12.

LCM(3, 6, 12) = 12.

In this case, the LCM (12) is equal to one of the numbers (12) in the set {3, 6, 12}.

From these examples, we see that it is indeed possible for the LCM of a set of numbers to be equal to one of the numbers in the set.

This happens when the largest number in the set is a multiple of all the other numbers in the set. If $\max(a_1, ..., a_n)$ is divisible by every $a_i$, then $\max(a_1, ..., a_n)$ is a common multiple. Since any common multiple must be greater than or equal to $\max(a_1, ..., a_n)$, the smallest common multiple must be $\max(a_1, ..., a_n)$.

The statement says that this "may be" the case, which means it is possible for the LCM to be one of the numbers. Our examples demonstrate that this is possible.

Answer:

The statement is True.

Given:

Two or more numbers and their HCF.

To Check:

The truth value of the statement: "HCF of two or more numbers may be one of the numbers".

Solution:

Let's consider a set of two or more positive integers: $\{a_1, a_2, ..., a_n\}$, where $n \ge 2$.

The Highest Common Factor (HCF), also known as the Greatest Common Divisor (GCD), of these numbers, denoted as HCF$(a_1, ..., a_n)$, is the largest positive integer that divides each number in the set.

By definition, for any number $a_i$ in the set, HCF$(a_1, ..., a_n)$ must divide $a_i$.

We need to check if it is possible for the HCF to be equal to one of the numbers in the set.

Let's consider some examples:

Example 1: Consider the numbers 4 and 8.

Factors of 4: 1, 2, 4

Factors of 8: 1, 2, 4, 8

The common factors are 1, 2, 4. The highest common factor is 4.

HCF(4, 8) = 4.

In this case, the HCF (4) is equal to one of the numbers (4) in the set {4, 8}.

Example 2: Consider the numbers 6 and 12.

Factors of 6: 1, 2, 3, 6

Factors of 12: 1, 2, 3, 4, 6, 12

The common factors are 1, 2, 3, 6. The highest common factor is 6.

HCF(6, 12) = 6.

In this case, the HCF (6) is equal to one of the numbers (6) in the set {6, 12}.

Example 3: Consider the numbers 3, 6, and 12.

Factors of 3: 1, 3

Factors of 6: 1, 2, 3, 6

Factors of 12: 1, 2, 3, 4, 6, 12

The common factors are 1, 3. The highest common factor is 3.

HCF(3, 6, 12) = 3.

In this case, the HCF (3) is equal to one of the numbers (3) in the set {3, 6, 12}.

From these examples, we see that it is indeed possible for the HCF of a set of numbers to be equal to one of the numbers in the set.

This happens when the smallest number in the set is a divisor of all the other numbers in the set. If $\min(a_1, ..., a_n)$ divides every $a_i$, then $\min(a_1, ..., a_n)$ is a common divisor. Since any common divisor must be less than or equal to HCF$(a_1, ..., a_n)$, and $\min(a_1, ..., a_n)$ is the largest possible value for a common divisor that is also one of the numbers, the HCF must be $\min(a_1, ..., a_n)$.

The statement says that this "may be" the case, which means it is possible for the HCF to be one of the numbers. Our examples demonstrate that this is possible.

Answer:

The statement is True.

Given:

Every whole number.

To Check:

The truth value of the statement: "Every whole number is the successor of another whole number".

Solution:

Whole numbers are the set of non-negative integers: $W = \{0, 1, 2, 3, ...\}$.

The successor of a whole number $n$ is the whole number that comes immediately after it. It is defined as $n+1$.

The statement claims that for every whole number, say $m \in W$, there exists another whole number, say $n \in W$, such that $m$ is the successor of $n$.

This means, for every $m \in W$, there exists $n \in W$ such that:

Let's examine the whole numbers one by one.

• Consider the whole number 1. Is 1 the successor of some whole number $n$? $1 = n + 1 \implies n = 1 - 1 = 0$. Yes, $n=0$ is a whole number. So, 1 is the successor of 0.
• Consider the whole number 2. Is 2 the successor of some whole number $n$? $2 = n + 1 \implies n = 2 - 1 = 1$. Yes, $n=1$ is a whole number. So, 2 is the successor of 1.
• Consider the whole number 3. Is 3 the successor of some whole number $n$? $3 = n + 1 \implies n = 3 - 1 = 2$. Yes, $n=2$ is a whole number. So, 3 is the successor of 2.

For any positive whole number $m > 0$, we can find $n = m-1$. Since $m > 0$, $m-1 \ge 0$. As long as $m$ is an integer, $m-1$ is also an integer. So, for any $m \in \{1, 2, 3, ...\}$, $n = m-1$ is a whole number, and $m$ is the successor of $n$.

Now, consider the whole number 0. Is 0 the successor of some whole number $n$? We need to find a whole number $n$ such that:

Solving for $n$, we get $n = 0 - 1 = -1$.

Is $n=-1$ a whole number? No, the set of whole numbers is $\{0, 1, 2, 3, ...\}$. Negative integers are not whole numbers.

So, there is no whole number $n$ such that 0 is the successor of $n$.

The number 0 is a whole number, but it is not the successor of another whole number.

The statement claims that every whole number is the successor of another whole number. This is false because the whole number 0 is not the successor of any whole number.

Answer:

The statement is False.

Given:

Sum and product of two whole numbers.

To Check:

The truth value of the statement: "Sum of two whole numbers is always less than their product".

Solution:

Let the two whole numbers be $a$ and $b$, where $a, b \in \{0, 1, 2, 3, ...\}$.

The sum is $a+b$ and the product is $a \times b$ or $ab$.

The statement claims that for all whole numbers $a$ and $b$, the following inequality is true:

To check if this statement is always true, we can test with some examples of whole numbers. If we find even one case where the inequality $a+b < ab$ does not hold (i.e., $a+b \ge ab$), then the statement is false.

Example 1: Let $a=0$ and $b=5$.

Sum $= a + b = 0 + 5 = 5$.

Product $= a \times b = 0 \times 5 = 0$.

Is $5 < 0$? No, $5 > 0$. In this case, the sum is greater than the product ($5 \ge 0$).

The statement $a+b < ab$ is false for $a=0, b=5$.

Example 2: Let $a=1$ and $b=1$.

Sum $= a + b = 1 + 1 = 2$.

Product $= a \times b = 1 \times 1 = 1$.

Is $2 < 1$? No, $2 > 1$. In this case, the sum is greater than the product ($2 \ge 1$).

The statement $a+b < ab$ is false for $a=1, b=1$.

Example 3: Let $a=1$ and $b=2$.

Sum $= a + b = 1 + 2 = 3$.

Product $= a \times b = 1 \times 2 = 2$.

Is $3 < 2$? No, $3 > 2$. In this case, the sum is greater than the product ($3 \ge 2$).

The statement $a+b < ab$ is false for $a=1, b=2$.

Example 4: Let $a=2$ and $b=2$.

Sum $= a + b = 2 + 2 = 4$.

Product $= a \times b = 2 \times 2 = 4$.

Is $4 < 4$? No, $4 = 4$. In this case, the sum is equal to the product ($4 \ge 4$).

The statement $a+b < ab$ is false for $a=2, b=2$.

We have found several counterexamples where the sum of two whole numbers is not strictly less than their product. The sum can be greater than or equal to the product in certain cases (e.g., when one number is 0 or 1, or when both numbers are 2).

The statement claims that the sum is *always* less than the product. Since we have found cases where this is not true, the statement is false.

The inequality $a+b < ab$ holds true for most pairs of larger whole numbers (e.g., $a=3, b=4$, sum=7, product=12; $7 < 12$). However, "always" requires it to be true for *all* pairs of whole numbers.

Answer:

The statement is False.

Given:

Two distinct whole numbers. Their sum is odd.

To Check:

The truth value of the statement: "then their difference also must be odd".

Solution:

Let the two distinct whole numbers be $a$ and $b$. Since they are distinct whole numbers, $a \ne b$, and $a, b \in \{0, 1, 2, 3, ...\}$.

The given condition is that the sum $a+b$ is odd.

The sum of two integers is odd if and only if one of the integers is even and the other is odd.

There are two possible scenarios for the parity of $a$ and $b$ for their sum to be odd:

Scenario 1: $a$ is even and $b$ is odd.

Scenario 2: $a$ is odd and $b$ is even.

Note that since $a$ and $b$ are distinct, both scenarios are possible (e.g., 2 and 3 in scenario 1, 3 and 2 in scenario 2). The order does not matter for the sum, and for the difference, we can consider $|a-b|$.

Let's analyze the difference between $a$ and $b$ in these two scenarios.

Consider the difference $a-b$ (or $|a-b|$ since we are interested in its parity).

If $a$ is even, $a$ can be written as $2m$ for some whole number $m$.

If $b$ is odd, $b$ can be written as $2n+1$ for some whole number $n$.

Scenario 1: $a$ is even, $b$ is odd.

Difference $= a - b = 2m - (2n + 1) = 2m - 2n - 1 = 2(m - n) - 1$.

Let $k = m-n$. If $m-n$ is an integer, $2k$ is an even integer. $2k - 1$ is an odd integer.

The difference $a-b$ is odd.

The absolute difference $|a-b|$ will also be odd (since the negative of an odd number is odd, and $|-odd| = odd$).

Scenario 2: $a$ is odd, $b$ is even.

Difference $= a - b = (2m + 1) - 2n = 2m - 2n + 1 = 2(m - n) + 1$.

Let $k = m-n$. If $m-n$ is an integer, $2k$ is an even integer. $2k + 1$ is an odd integer.

The difference $a-b$ is odd.

The absolute difference $|a-b|$ will also be odd.

In both scenarios, where the sum of two numbers is odd (meaning one is even and the other is odd), their difference is always odd.

Let's test with examples of distinct whole numbers:

• Numbers: 2 and 3. Sum = $2+3=5$ (odd). Difference = $|2-3| = |-1| = 1$ (odd). The statement holds.
• Numbers: 0 and 5. Sum = $0+5=5$ (odd). Difference = $|0-5| = |-5| = 5$ (odd). The statement holds.
• Numbers: 4 and 7. Sum = $4+7=11$ (odd). Difference = $|4-7| = |-3| = 3$ (odd). The statement holds.

The parity of the sum and difference of two numbers is always the same. That is, $(a+b)$ and $(a-b)$ have the same parity.

$(a+b) + (a-b) = 2a$ (an even number)

$(a+b) - (a-b) = 2b$ (an even number)

If the sum $(a+b)$ is odd, then $(a+b) + (a-b) =$ an even number. This can only happen if $(a-b)$ is also odd (Odd + Odd = Even).

If the sum $(a+b)$ is odd, then $(a+b) - (a-b) =$ an even number. This can only happen if $(a-b)$ is also odd (Odd - Odd = Even).

Therefore, if the sum of two numbers is odd, their difference must also be odd.

The condition that the numbers are distinct is important because if $a=b$, then $a+b = 2a$ (even) and $a-b=0$ (even). So if the sum is odd, the numbers must be distinct anyway.

Answer:

The statement is True.

Given:

Any two consecutive numbers.

To Check:

The truth value of the statement: "Any two consecutive numbers are coprime".

Solution:

Let the two consecutive whole numbers be $n$ and $n+1$, where $n$ is a whole number ($n \in \{0, 1, 2, 3, ...\}$).

Two numbers are said to be coprime (or relatively prime) if their Highest Common Factor (HCF) or Greatest Common Divisor (GCD) is 1.

The statement claims that for any whole number $n \ge 0$, HCF$(n, n+1) = 1$.

Let's consider a positive integer $d$ that is a common divisor of $n$ and $n+1$.

If $d$ is a common divisor of $n$ and $n+1$, then $d$ must divide $n$ exactly, and $d$ must divide $n+1$ exactly.

According to the property of divisibility, if a number $d$ divides two numbers, then $d$ must also divide their difference.

Consider the difference between the two consecutive numbers: $(n+1) - n$.

Since $d$ divides both $n$ and $n+1$, it must divide their difference, which is 1.

The only positive integer that divides 1 is 1 itself.

Therefore, the only positive common divisor of any two consecutive integers $n$ and $n+1$ is 1.

Since the greatest positive common divisor is 1, the HCF$(n, n+1) = 1$ for any integer $n$.

This holds true for whole numbers where $n \ge 0$.

Let's test the first few pairs of consecutive whole numbers:

• Numbers: 0 and 1. HCF(0, 1) = 1. They are coprime.
• Numbers: 1 and 2. HCF(1, 2) = 1. They are coprime.
• Numbers: 2 and 3. HCF(2, 3) = 1. They are coprime.
• Numbers: 3 and 4. HCF(3, 4) = 1. They are coprime.
• Numbers: 10 and 11. HCF(10, 11) = 1. They are coprime.

The proof using the difference $(n+1)-n=1$ confirms that the only common positive divisor is always 1 for any consecutive integers.

Thus, any two consecutive whole numbers are coprime.

Answer:

The statement is True.

Given:

Two numbers. Their HCF is one of the numbers.

To Check:

The truth value of the statement: "then their LCM is the other number".

Solution:

Let the two positive numbers be $a$ and $b$.

The given condition is that the HCF of $a$ and $b$ is one of the numbers. This means either HCF$(a, b) = a$ or HCF$(a, b) = b$.

By the definition of HCF, HCF$(a, b)$ is the largest positive integer that divides both $a$ and $b$.

If HCF$(a, b) = a$, this means $a$ divides both $a$ and $b$. $a$ always divides $a$. For $a$ to divide $b$, $b$ must be a multiple of $a$. So, $b = a \times k$ for some positive integer $k$. Since $a$ is the HCF, $a$ must be the largest number that divides both. If $b$ is a multiple of $a$ ($b=ak$), the common divisors are the divisors of $a$. The largest such divisor is $a$. So, if $b$ is a multiple of $a$, HCF$(a, b) = a$.

Similarly, if HCF$(a, b) = b$, this means $b$ divides both $a$ and $b$. $b$ always divides $b$. For $b$ to divide $a$, $a$ must be a multiple of $b$. So, $a = b \times k$ for some positive integer $k$. If $a$ is a multiple of $b$, HCF$(a, b) = b$.

So, the condition "HCF of two numbers is one of the numbers" is equivalent to saying that one number is a multiple of the other.

Let's assume, without loss of generality, that HCF$(a, b) = a$. This implies that $b$ is a multiple of $a$. So, $b = ak$ for some positive integer $k$. Since $a$ is the HCF, the largest common divisor is $a$. For $b$ to be a multiple of $a$ and $a$ to be the HCF, $k$ must be an integer. For example, HCF(3, 6) = 3. Here $b=6, a=3$, $6=3 \times 2$, $k=2$. HCF(4, 4) = 4. Here $b=4, a=4$, $4=4 \times 1$, $k=1$.

We are asked if the LCM of $a$ and $b$ is the other number. In this case, if HCF$(a, b) = a$, is LCM$(a, b) = b$?

Let's find the LCM of $a$ and $b$ when $b$ is a multiple of $a$, i.e., $b = ak$ for some positive integer $k$. The multiples of $a$ are $a, 2a, 3a, ..., ka, ...$ and the multiples of $b$ are $b, 2b, 3b, ...$. Since $b=ak$, the multiples of $b$ are $ak, 2ak, 3ak, ...$.

We are looking for the smallest positive common multiple of $a$ and $b$. Since $b=ak$, $b$ is a multiple of $a$. The multiples of $a$ are $\{a, 2a, 3a, ..., (k-1)a, ka, (k+1)a, ...\}$. The multiples of $b$ are $\{b, 2b, 3b, ...\} = \{ak, 2ak, 3ak, ...\}$.

The number $b=ak$ is a multiple of $a$ (since $k$ is an integer) and is a multiple of $b$ (since $b = 1 \times b$). So, $b$ is a common multiple of $a$ and $b$.

Is $b$ the lowest common multiple? Since $b$ is a multiple of $a$, any common multiple of $a$ and $b$ must also be a multiple of $b$. The smallest positive multiple of $b$ is $b$ itself.

Therefore, if $b$ is a multiple of $a$, then LCM$(a, b) = b$.

So, if HCF$(a, b) = a$ (which implies $b$ is a multiple of $a$), then LCM$(a, b) = b$ (which is the other number). This holds true.

Similarly, if HCF$(a, b) = b$ (which implies $a$ is a multiple of $b$), then LCM$(a, b) = a$ (which is the other number). This also holds true.

The statement "If the HCF of two numbers is one of the numbers, then their LCM is the other number" is true for positive integers.

Let's check with examples used in the previous question (Q90):

Example 1: Numbers 4 and 8. HCF(4, 8) = 4 (one of the numbers). LCM(4, 8) = 8 (the other number). The statement holds.

Example 2: Numbers 6 and 12. HCF(6, 12) = 6 (one of the numbers). LCM(6, 12) = 12 (the other number). The statement holds.

Example 3: Numbers 3 and 12. HCF(3, 12) = 3 (one of the numbers). Multiples of 3: 3, 6, 9, 12, ... Multiples of 12: 12, 24, ... LCM(3, 12) = 12 (the other number). The statement holds.

This property is a direct consequence of the relationship $a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$.

If HCF$(a, b) = a$, then $ab = a \times \text{LCM}(a, b)$. Dividing by $a$ (assuming $a \ne 0$), we get $b = \text{LCM}(a, b)$.

If HCF$(a, b) = b$, then $ab = b \times \text{LCM}(a, b)$. Dividing by $b$ (assuming $b \ne 0$), we get $a = \text{LCM}(a, b)$.

The statement holds true for non-zero numbers. If one number is 0, the HCF is the other number (if non-zero), and the LCM is typically defined as 0. E.g., HCF(5, 0) = 5, LCM(5, 0) = 0. Here HCF is one number (5), but LCM (0) is not the other number (5). However, the question seems to imply positive numbers as is common for HCF/LCM problems unless specified otherwise, or the case with 0 is considered an edge case where the standard rules might not apply in the same way. Assuming positive numbers, the statement is true.

Answer:

The statement is True (assuming positive numbers).

Given:

Two numbers and their HCF.

To Check:

The truth value of the statement: "The HCF of two numbers is smaller than the smaller of the numbers".

Solution:

Let the two positive numbers be $a$ and $b$. Assume, without loss of generality, that $a \le b$. The smaller of the two numbers is $a$.

Let $h = \text{HCF}(a, b)$.

By definition, the HCF is the largest positive integer that divides both $a$ and $b$.

Since $h$ divides $a$, it must be that $h$ is less than or equal to $a$.

The statement claims that the HCF is smaller than the smaller of the numbers, which means $h < a$ (since $a$ is the smaller or equal number).

We need to check if there are any cases where $h$ is not strictly smaller than $a$, i.e., where $h = a$.

Consider the case where the smaller number $a$ divides the larger number $b$.

If $a$ divides $b$, then $a$ is a common divisor of $a$ and $b$ (since $a$ divides $a$ and $a$ divides $b$).

Since $a$ divides $a$, any common divisor of $a$ and $b$ must also be a divisor of $a$.

The largest divisor of $a$ is $a$ itself.

Therefore, if $a$ divides $b$, the largest common divisor of $a$ and $b$ is $a$.

In this case, HCF$(a, b) = a$.

Since we assumed $a \le b$, $a$ is the smaller (or equal) number. The HCF is equal to the smaller number $a$.

This contradicts the statement that HCF is smaller than the smaller number ($h < a$).

Let's consider an example:

Numbers: 4 and 8. The smaller number is 4. HCF(4, 8) = 4.

Is HCF(4) smaller than the smaller number (4)? Is $4 < 4$? No, $4 = 4$.

This example shows that the HCF is not always strictly smaller than the smaller number; it can be equal to the smaller number.

Another example:

Numbers: 7 and 14. The smaller number is 7. HCF(7, 14) = 7.

Is HCF(7) smaller than the smaller number (7)? Is $7 < 7$? No, $7 = 7$.

This also contradicts the statement.

The HCF of two numbers is always less than or equal to the smaller of the two numbers.

The statement requires strict inequality ($<$), which is not always true.

Answer:

The statement is False.

Given:

Two numbers and their LCM.

To Check:

The truth value of the statement: "The LCM of two numbers is greater than the larger of the numbers".

Solution:

Let the two positive numbers be $a$ and $b$. Assume, without loss of generality, that $a \le b$. The larger of the two numbers is $b$.

Let $l = \text{LCM}(a, b)$.

By definition, the LCM is the smallest positive integer that is a multiple of both $a$ and $b>.

Since $l$ is a multiple of $b$, it must be that $l$ is greater than or equal to $b$.

The statement claims that the LCM is greater than the larger of the numbers, which means $l > b$ (since $b$ is the larger or equal number).

We need to check if there are any cases where $l$ is not strictly greater than $b$, i.e., where $l = b$.

Consider the case where the smaller number $a$ divides the larger number $b$.

If $a$ divides $b$, then $b$ is a multiple of $a$. So, $b = ak$ for some positive integer $k$.

The multiples of $a$ are $a, 2a, ..., ka, ...$ and the multiples of $b$ are $b, 2b, ...$.

Since $b = ak$, the number $b$ is present in the list of multiples of $a$. Also, $b$ is trivially a multiple of $b$.

So, $b$ is a common multiple of $a$ and $b$.

Since any common multiple of $a$ and $b$ must be a multiple of their LCM, and $b$ is a common multiple, LCM$(a, b)$ must divide $b$. Also, by definition, LCM$(a, b)$ is a multiple of $b$, so LCM$(a, b) \ge b$.

Given that $b$ is a multiple of $a$, the multiples of $a$ are $\{a, 2a, ..., b, (k+1)a, ...\}$. The multiples of $b$ are $\{b, 2b, ...\}$. The smallest positive number appearing in both lists is $b$.

Therefore, if $a$ divides $b$, then LCM$(a, b) = b$.

Since we assumed $a \le b$, $b$ is the larger (or equal) number. The LCM is equal to the larger number $b$.

This contradicts the statement that LCM is greater than the larger number ($l > b$).

Let's consider an example:

Numbers: 4 and 8. The larger number is 8. LCM(4, 8) = 8.

Is LCM(8) greater than the larger number (8)? Is $8 > 8$? No, $8 = 8$.

This example shows that the LCM is not always strictly greater than the larger number; it can be equal to the larger number.

Another example:

Numbers: 7 and 14. The larger number is 14. LCM(7, 14) = 14.

Is LCM(14) greater than the larger number (14)? Is $14 > 14$? No, $14 = 14$.

This also contradicts the statement.

The LCM of two positive numbers is always greater than or equal to the larger of the two numbers.

The statement requires strict inequality ($>$), which is not always true.

Note: If the numbers are coprime (like 3 and 5), HCF(3, 5) = 1, LCM(3, 5) = 15. The larger number is 5. $15 > 5$, so the statement holds in this case. But it does not hold "always".

Answer:

The statement is False.

Given:

Two coprime numbers.

To Check:

The truth value of the statement: "The LCM of two coprime numbers is equal to the product of the numbers".

Solution:

Let the two positive numbers be $a$ and $b$.

The given condition is that $a$ and $b$ are coprime. This means their Highest Common Factor (HCF) is 1.

We need to check if their Lowest Common Multiple (LCM) is equal to their product $a \times b$.

We know the fundamental relationship between the HCF and LCM of two positive numbers $a$ and $b$:

Given that the numbers $a$ and $b$ are coprime, their HCF is 1.

Substitute HCF$(a, b) = 1$ into the formula:

This equation shows that if the HCF of two numbers is 1, then their product is equal to their LCM.

This confirms the statement.

Let's consider examples of coprime numbers:

Example 1: Numbers 3 and 5.

HCF(3, 5) = 1. They are coprime.

Product $= 3 \times 5 = 15$.

Multiples of 3: 3, 6, 9, 12, 15, 18, ...

Multiples of 5: 5, 10, 15, 20, ...

LCM(3, 5) = 15.

Here, LCM = 15 and product = 15. LCM = Product. The statement holds.

Example 2: Numbers 4 and 7.

HCF(4, 7) = 1. They are coprime.

Product $= 4 \times 7 = 28$.

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, ...

Multiples of 7: 7, 14, 21, 28, ...

LCM(4, 7) = 28.

Here, LCM = 28 and product = 28. LCM = Product. The statement holds.

The statement is a direct consequence of the relationship between HCF, LCM, and the product of two numbers. It holds true for any pair of positive coprime integers.

Answer:

The statement is True.

(a) We know that 1 million is equal to 10 lakh, and 1 crore is equal to 100 lakh.

Therefore, 10 million = $10 \times 10$ lakh = 100 lakh.

Since 100 lakh = 1 crore, 10 million = 1 crore.

(b) We know that 1 million is equal to 10 lakh.

Therefore, 10 lakh = 1 million.

(a) We know that 1 metre = 100 centimetres.

Also, 1 centimetre = 10 millimetres.

Therefore, 1 metre = $100 \times 10$ millimetres = 1000 millimetres.

So, 1 metre = 1000 millimetres.

(b) We know that 1 centimetre = 10 millimetres.

So, 1 centimetre = 10 millimetres.

(c) We know that 1 kilometre = 1000 metres.

Also, 1 metre = 1000 millimetres.

Therefore, 1 kilometre = $1000 \times 1000$ millimetres = 1,000,000 millimetres.

So, 1 kilometre = 1,000,000 millimetres.

(a) We know that 1 gram is equal to 1000 milligrams.

So, 1 gram = 1000 milligrams.

(b) We know that 1 litre is equal to 1000 millilitres.

So, 1 litre = 1000 millilitres.

(c) We know that 1 kilogram = 1000 grams.

Also, 1 gram = 1000 milligrams.

Therefore, 1 kilogram = $1000 \times 1000$ milligrams = 1,000,000 milligrams.

So, 1 kilogram = 1,000,000 milligrams.

We know that 1 lakh is equal to 100 thousands.

Therefore, 100 thousands = 1 lakh.

Given height = 1m 65cm.

We need to convert this height into millimetres.

We know that 1 metre = 1000 millimetres.

So, 1m = 1000 mm.

We also know that 1 centimetre = 10 millimetres.

So, 65cm = $65 \times 10$ millimetres = 650 millimetres.

Total height in millimetres = Height in metres (in mm) + Height in centimetres (in mm).

Total height = $1000 \text{ mm} + 650 \text{ mm} = 1650 \text{ mm}$.

His height in millimetres is 1650.

Given length of river Narmada = 1290 km.

We need to convert this length into metres.

We know that 1 kilometre = 1000 metres.

Therefore, 1290 km = $1290 \times 1000$ metres = 1,290,000 metres.

Its length in metres is 1,290,000.

Given distance between Srinagar and Leh = 422 km.

We need to convert this distance into metres.

We know that 1 kilometre = 1000 metres.

Therefore, 422 km = $422 \times 1000$ metres = 422,000 metres.

The same distance in metres is 422,000.

The greatest number made by five different non-zero digits uses the digits 9, 8, 7, 6, and 5 arranged in descending order.

The greatest number is 98765.

Reversing the order of digits of 98765 gives 56789.

This new number, 56789, is formed by the same five digits (5, 6, 7, 8, 9) but arranged in ascending order.

Arranging digits in ascending order forms the smallest number possible using those specific digits.

Therefore, the new number is the smallest number of five digits formed by these digits.

The new number is the smallest number of five digits (using the same digits).

We know that ten lakh is written as 10,00,000.

The number that is 1 less than 10,00,000 is $10,00,000 - 1 = 9,99,999$.

The number 9,99,999 has 6 digits and is the greatest 6-digit number.

So, by adding 1 to the greatest 6-digit number (9,99,999), we get ten lakh (10,00,000).

By adding 1 to the greatest six digit number, we get ten lakh.

The number given in words is five crore twenty three lakh seventy eight thousand four hundred one.

We can break this down by place value in the Indian System:

Five crore: 5,00,00,000

Twenty three lakh: 23,00,000

Seventy eight thousand: 78,000

Four hundred one: 401

Adding these values together:

$5,00,00,000 + 23,00,000 + 78,000 + 401 = 5,23,78,401$.

Using commas in the Indian System of Numeration, the number is written as 5,23,78,401.

The number can be written, using commas, in the Indian System of Numeration as 5,23,78,401.

In Roman Numeration, specific rules govern subtraction. The symbol X can be subtracted from symbols of larger value.

According to the standard rules of Roman numerals:

The symbol I can be subtracted from V and X.

The symbol X can be subtracted from L and C.

The symbol C can be subtracted from D and M.

The question states that the symbol X can be subtracted from _____, M and C only.

Based on the standard rules, X can be subtracted from L and C.

The list given in the question includes C and M, and a blank space.

Comparing the standard list of symbols from which X can be subtracted ({L, C}) with the list provided in the question ({blank, M, C}), the most likely symbol for the blank is L, as it is the other standard symbol from which X can be subtracted.

While the inclusion of M in the question's list and the word "only" make the complete statement inconsistent with the standard rules (X is not standardly subtracted from M), the task is to fill the blank based on the context provided, which appears to be referring to the standard subtraction rules where X is subtracted from L and C.

Therefore, the blank should be L.

In Roman Numeration, the symbol X can be subtracted from L, M and C only.

To write the number 66 in Roman numerals, we decompose it based on the standard Roman numeral values:

$66 = 50 + 10 + 5 + 1$

The Roman numeral for 50 is L.

The Roman numeral for 10 is X.

The Roman numeral for 5 is V.

The Roman numeral for 1 is I.

Combining these symbols in descending order of their values:

$50 \to \text{L}$

$10 \to \text{X}$

$5 \to \text{V}$

$1 \to \text{I}$

Putting them together, we get LXVI.

The number 66 in Roman numerals is LXVI.

Given population is 2,538,473.

We need to round this number to the nearest thousands.

The thousands digit is 8.

The digit to the right of the thousands digit (in the hundreds place) is 4.

Since the digit 4 is less than 5, we keep the thousands digit as it is (8) and replace all digits to its right with zeros.

The digits to the right are 4, 7, and 3.

Replacing these digits with zeros, we get 2,538,000.

Rounded off to nearest thousands, the population was 2,538,000.

Whole numbers are the set of non-negative integers. They start from 0 and go upwards infinitely (0, 1, 2, 3, ...).

The smallest number in this set is the first number listed.

The smallest whole number is 0.

The successor of a number is the number that comes immediately after it. It is obtained by adding 1 to the given number.

Successor of 106159 = $106159 + 1$.

$106159 + 1 = 106160$.

Successor of 106159 is 106160.

The predecessor of a number is the number that comes immediately before it. It is obtained by subtracting 1 from the given number.

Predecessor of 100000 = $100000 - 1$.

$100000 - 1 = 99999$.

Predecessor of 100000 is 99999.

If 400 is the predecessor of a number, it means that number is 1 more than 400.

The number = $400 + 1 = 401$.

400 is the predecessor of 401.

The largest 3-digit number is 999.

The successor of 999 is the number immediately after it, which is $999 + 1 = 1000$.

1000 is the smallest 4-digit number.

1000 is the successor of the largest 3 digit number.

Let $a$ be a whole number.

Subtracting 0 from $a$ means performing the operation $a - 0$.

Any number minus 0 is the number itself.

$a - 0 = a$

So, the result is the whole number itself.

If 0 is subtracted from a whole number, then the result is the number itself .

The smallest 6-digit natural number is 100000.

We are looking for the smallest 6-digit natural number that ends in 5.

The smallest 6-digit numbers are 100000, 100001, 100002, 100003, 100004, 100005, ...

Among these numbers, the first one that ends in 5 is 100005.

Alternatively, start with the smallest 6-digit number (100000) and find the nearest number greater than or equal to it that ends in 5.

100000 ends in 0.

To make it end in 5, we need to increase it. The next multiple of 10 ends in 0, the one after that ends in 5.

100000 + 5 = 100005.

This is a 6-digit number and ends in 5. Any smaller 6-digit number would be less than 100005.

The smallest 6 digit natural number ending in 5 is 100005.

The set of whole numbers is $\{0, 1, 2, 3, ...\}$.

A set is closed under an operation if, when you perform the operation on any two elements from the set, the result is also an element of the set.

Consider the basic operations:

Addition: For any two whole numbers $a$ and $b$, the sum $a+b$ is always a whole number. For example, $5+8=13$, $0+7=7$. Both 13 and 7 are whole numbers. So, whole numbers are closed under addition.

Subtraction: For any two whole numbers $a$ and $b$, the difference $a-b$ is not always a whole number. For example, $3-5=-2$, and -2 is not a whole number. So, whole numbers are not closed under subtraction.

Multiplication: For any two whole numbers $a$ and $b$, the product $a \times b$ is always a whole number. For example, $5 \times 8=40$, $0 \times 7=0$. Both 40 and 0 are whole numbers. So, whole numbers are closed under multiplication.

Division: For any two whole numbers $a$ and $b$ (where $b \neq 0$), the quotient $a \div b$ is not always a whole number. For example, $3 \div 5 = \frac{3}{5}$, which is not a whole number. So, whole numbers are not closed under division.

Thus, whole numbers are closed under addition and multiplication.

Whole numbers are closed under addition and under multiplication.

The set of natural numbers is $\{1, 2, 3, ...\}$.

Let's examine the basic arithmetic operations to see if the result is always a natural number when we operate on two natural numbers.

Addition: For any two natural numbers $a$ and $b$, their sum $a+b$ is always a natural number. For example, $2+5=7$, which is a natural number. $10+1=11$, which is a natural number. So, natural numbers are closed under addition.

Subtraction: For any two natural numbers $a$ and $b$, their difference $a-b$ is not always a natural number. For example, $3-7=-4$, which is not a natural number. So, natural numbers are not closed under subtraction.

Multiplication: For any two natural numbers $a$ and $b$, their product $a \times b$ is always a natural number. For example, $2 \times 5 = 10$, which is a natural number. $10 \times 1 = 10$, which is a natural number. So, natural numbers are closed under multiplication.

Division: For any two natural numbers $a$ and $b$ (where $b \neq 0$), their quotient $a \div b$ is not always a natural number. For example, $2 \div 5 = \frac{2}{5}$, which is not a natural number. So, natural numbers are not closed under division.

Thus, natural numbers are closed under addition and multiplication.

Natural numbers are closed under addition and under multiplication.

In mathematics, division by zero is an undefined operation.

If we try to divide a non-zero number by zero, the result is undefined.

If we try to divide zero by zero, the result is also undefined (it is considered an indeterminate form).

Since 0 is a whole number, the division of any whole number (including 0) by 0 is not defined.

Division of a whole number by zero is not defined.

The distributive property in mathematics states that multiplication distributes over addition and subtraction.

For any three whole numbers $a$, $b$, and $c$:

Distributivity over addition: $a \times (b + c) = (a \times b) + (a \times c)$.

Let's check with whole numbers:

Let $a=2$, $b=3$, $c=4$.

$2 \times (3 + 4) = 2 \times 7 = 14$.

$(2 \times 3) + (2 \times 4) = 6 + 8 = 14$.

Since $14 = 14$, multiplication is distributive over addition for whole numbers.

Distributivity over subtraction: $a \times (b - c) = (a \times b) - (a \times c)$, provided $b \geq c$ so that $b-c$ is a whole number.

Let's check with whole numbers (where $b \geq c$):

Let $a=2$, $b=4$, $c=3$.

$2 \times (4 - 3) = 2 \times 1 = 2$.

$(2 \times 4) - (2 \times 3) = 8 - 6 = 2$.

Since $2 = 2$, multiplication is distributive over subtraction for whole numbers (when the subtraction results in a whole number).

The most commonly referred-to distributive property is over addition.

Multiplication is distributive over addition for whole numbers.

The given equation is $2395 \times \text{_____} = 6195 \times 2395$.

This equation demonstrates the commutative property of multiplication.

The commutative property of multiplication states that for any two numbers $a$ and $b$, the order of multiplication does not affect the product. That is, $a \times b = b \times a$.

In the given equation, if we let $a = 2395$ and $b = 6195$, the right side is $b \times a$.

So, we have $a \times \text{_____} = b \times a$.

According to the commutative property, this equality holds when the blank is filled with $b$.

Therefore, the number in the blank is 6195.

$2395 \times 6195 = 6195 \times 2395$.

The given equation is $1001 \times 2002 = 1001 \times (1001 + \text{_____})$.

This equation demonstrates the distributive property of multiplication over addition, which states $a \times (b + c) = a \times b + a \times c$.

In the given equation, the left side is $1001 \times 2002$.

The right side has the factor 1001 multiplied by an expression in parentheses, $(1001 + \text{_____})$.

For the equality to hold, the expression inside the parentheses must be equal to 2002.

So, we have the equation:

To find the value of the blank, we subtract 1001 from 2002.

Thus, the number in the blank is 1001.

1001 × 2002 = 1001 × (1001 + 1001 ).

The given expression is $10001 \times 0$.

According to the property of multiplication by zero, any number multiplied by zero is equal to zero.

$a \times 0 = 0$ for any number $a$.

In this case, $a = 10001$.

So, $10001 \times 0 = 0$.

10001 × 0 = 0.

The given equation is $2916 \times \text{_____} = 0$.

We know that for the product of two numbers to be zero, at least one of the numbers must be zero.

In this equation, one factor is 2916, which is not zero.

Therefore, the other factor (the number in the blank) must be zero for the product to be zero.

$2916 \times 0 = 0$.

2916 × 0 = 0.

The given equation is $9128 \times \text{_____} = 9128$.

This equation demonstrates the property of the multiplicative identity for whole numbers.

The multiplicative identity for whole numbers is 1, because any whole number multiplied by 1 results in the same whole number.

$a \times 1 = a$ for any whole number $a$.

In this case, $a = 9128$.

For the equation $9128 \times \text{_____} = 9128$ to be true, the number in the blank must be 1.

$9128 \times 1 = 9128$.

9128 × 1 = 9128.

The given equation is $125 + (68 + 17) = (125 + \text{_____}) + 17$.

This equation demonstrates the associative property of addition.

The associative property of addition states that for any three numbers $a$, $b$, and $c$, the way in which the numbers are grouped for addition does not affect the sum. That is, $a + (b + c) = (a + b) + c$.

In the given equation, let $a = 125$, $b = 68$, and $c = 17$.

The left side is in the form $a + (b + c)$.

The right side is in the form $(a + \text{_____}) + c$.

Comparing the two sides based on the associative property, the blank should be filled with $b$.

Therefore, the number in the blank is 68.

$125 + (68 + 17) = (125 + 68 ) + 17$.

The given expression is $8925 \times 1$.

This demonstrates the property of the multiplicative identity.

According to the multiplicative identity property, when any number is multiplied by 1, the result is the number itself.

$a \times 1 = a$ for any number $a$.

In this case, $a = 8925$.

So, $8925 \times 1 = 8925$.

8925 × 1 = 8925.

The given equation is $19 \times 12 + 19 = 19 \times (12 + \text{_____})$.

This equation demonstrates the distributive property of multiplication over addition in reverse.

The distributive property is $a \times (b + c) = (a \times b) + (a \times c)$.

Looking at the left side of the given equation, $19 \times 12 + 19$, we can rewrite the second term as $19 \times 1$ because 19 is the same as $19 \times 1$.

So, the left side becomes $19 \times 12 + 19 \times 1$.

Now, the equation is $19 \times 12 + 19 \times 1 = 19 \times (12 + \text{_____})$.

By the distributive property, we can factor out the common factor of 19 from the left side:

Comparing this with the right side of the given equation, $19 \times (12 + \text{_____})$, we see that the number in the blank must be 1.

19 × 12 + 19 = 19 × (12 + 1).

The given equation is $24 \times 35 = 24 \times 18 + 24 \times \text{_____}$.

This equation illustrates the distributive property of multiplication over addition.

The distributive property states that for numbers $a$, $b$, and $c$, $a \times (b + c) = (a \times b) + (a \times c)$.

Comparing the given equation with the distributive property, we can see that:

So, we have the equation $24 \times 35 = (24 \times 18) + (24 \times c)$, where $c$ is the number in the blank.

According to the distributive property, $24 \times (18 + c) = (24 \times 18) + (24 \times c)$.

Equating the left side of the given equation to this form:

For this equality to hold, the expression inside the parentheses must be equal:

To find the value of c, we subtract 18 from 35:

Therefore, the number in the blank is 17.

24 × 35 = 24 × 18 + 24 × 17.

The given equation is $32 \times (27 \times 19) = (32 \times \text{_____}) \times 19$.

This equation demonstrates the associative property of multiplication.

The associative property of multiplication states that for any three numbers $a$, $b$, and $c$, the way in which the numbers are grouped for multiplication does not affect the product. That is, $a \times (b \times c) = (a \times b) \times c$.

In the given equation, let $a = 32$, $b = 27$, and $c = 19$.

The left side is in the form $a \times (b \times c)$.

The right side is in the form $(a \times \text{_____}) \times c$.

Comparing the two sides based on the associative property, the blank should be filled with $b$.

Therefore, the number in the blank is 27.

32 × (27 × 19) = (32 × 27 ) × 19.

The given expression is $786 \times 3 + 786 \times 7$.

We can use the distributive property of multiplication over addition, which states that $a \times b + a \times c = a \times (b + c)$.

In this expression, we have $a = 786$, $b = 3$, and $c = 7$.

Applying the distributive property:

Now, we perform the addition inside the parentheses:

Substitute this back into the expression:

Perform the multiplication:

So, the value of the expression is 7860.

786 × 3 + 786 × 7 = 7860.

The given equation is $24 \times 25 = 24 \times \text{_____}$.

For the equality to be true, the expression on the left side must be equal to the expression on the right side.

The left side is $24 \times 25$.

The right side is $24 \times \text{_____}$.

Since 24 is a common factor on both sides, the number in the blank must be equal to 25 for the equality to hold.

Thus, the blank should be filled with 25.

24 × 25 = 24 × 25.

A factor of a number is a number that divides the given number exactly, with no remainder.

For example, the factors of 12 are 1, 2, 3, 4, 6, and 12.

Let's look at the relationship between the number (12) and its factors (1, 2, 3, 4, 6, 12).

12 is a multiple of 1 ($1 \times 12 = 12$).

12 is a multiple of 2 ($2 \times 6 = 12$).

12 is a multiple of 3 ($3 \times 4 = 12$).

12 is a multiple of 4 ($4 \times 3 = 12$).

12 is a multiple of 6 ($6 \times 2 = 12$).

12 is a multiple of 12 ($12 \times 1 = 12$).

In general, if $f$ is a factor of a number $n$, it means $n \div f$ is a whole number (or $n = f \times k$ for some whole number $k$).

This also means that $n$ is a multiple of $f$.

So, a number is a multiple of each of its factors.

A number is a multiple of each of its factor.

A factor of a number is a number that divides the given number exactly, with no remainder.

Let's consider a few numbers and their factors:

Factors of 5 are 1, 5.

Factors of 10 are 1, 2, 5, 10.

Factors of 12 are 1, 2, 3, 4, 6, 12.

Factors of any positive integer $n$ include 1, because $n \div 1 = n$, which is a whole number with a remainder of 0. So, 1 divides every positive integer exactly.

For the number 0, any non-zero number is a factor of 0 because $0 \div a = 0$ for any $a \neq 0$. However, 1 is also a factor of 0 because $0 \div 1 = 0$.

Thus, 1 is a factor of every whole number.

1 is a factor of every number.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's consider a few prime numbers and find their factors:

Prime number 2: The only positive integers that divide 2 exactly are 1 and 2. Factors are {1, 2}. Number of factors = 2.

Prime number 3: The only positive integers that divide 3 exactly are 1 and 3. Factors are {1, 3}. Number of factors = 2.

Prime number 5: The only positive integers that divide 5 exactly are 1 and 5. Factors are {1, 5}. Number of factors = 2.

Prime number 7: The only positive integers that divide 7 exactly are 1 and 7. Factors are {1, 7}. Number of factors = 2.

By definition, a prime number $p$ has exactly two distinct positive divisors: 1 and $p$ itself.

The number of factors of a prime number is two.

Let's consider an example to understand this property.

Consider the number 6.

The factors of 6 are the numbers that divide 6 exactly: 1, 2, 3, and 6.

The sum of the factors of 6 is $1 + 2 + 3 + 6 = 12$.

Twice the number 6 is $2 \times 6 = 12$.

Since the sum of the factors (12) is equal to twice the number (12), 6 satisfies the given condition.

Consider another example, the number 28.

The factors of 28 are 1, 2, 4, 7, 14, and 28.

The sum of the factors of 28 is $1 + 2 + 4 + 7 + 14 + 28 = 56$.

Twice the number 28 is $2 \times 28 = 56$.

Since the sum of the factors (56) is equal to twice the number (56), 28 also satisfies the given condition.

Numbers that have this property (the sum of all their factors is equal to twice the number) are called perfect numbers.

A number for which the sum of all its factors is equal to twice the number is called a perfect number.

We know that a prime number is a natural number greater than 1 that has exactly two factors: 1 and itself.

Natural numbers greater than 1 that are not prime are called composite numbers.

Let's consider a few examples of numbers greater than 1:

Number 2: Factors are {1, 2}. Number of factors = 2 (Prime).

Number 3: Factors are {1, 3}. Number of factors = 2 (Prime).

Number 4: Factors are {1, 2, 4}. Number of factors = 3 (More than two factors).

Number 5: Factors are {1, 5}. Number of factors = 2 (Prime).

Number 6: Factors are {1, 2, 3, 6}. Number of factors = 4 (More than two factors).

Numbers with more than two factors must have factors other than 1 and themselves. This is the definition of a composite number.

The numbers having more than two factors are called composite numbers.

Let's consider different types of numbers:

Even numbers: Integers that are divisible by 2 (..., -4, -2, 0, 2, 4, ...).

Odd numbers: Integers that are not divisible by 2 (..., -3, -1, 1, 3, 5, ...).

Prime numbers: Natural numbers greater than 1 with exactly two factors (1 and itself). Examples: 2, 3, 5, 7, 11, ...

Composite numbers: Natural numbers greater than 1 with more than two factors. Examples: 4, 6, 8, 9, 10, ...

The number 2 is an even number because it is divisible by 2.

The factors of 2 are 1 and 2. Since it has exactly two factors, it is a prime number.

Consider any other even number greater than 2. Any such number can be written as $2k$, where $k$ is an integer greater than 1 (since the number is > 2 and even). For example, 4 = 2 $\times$ 2, 6 = 2 $\times$ 3, 8 = 2 $\times$ 4, etc.

Any even number greater than 2 will have factors 1, 2, the number itself, and also $k$ (where $k > 1$). For $k > 1$, $k$ is a factor other than 1 and the number itself (unless $k$ equals the number, which only happens if the number is 2). Since $k>1$, if the number is even and greater than 2, $k$ will be a factor other than 1. If $k$ is not equal to the number itself (which is true for even numbers greater than 2), then the number has at least 3 factors (1, 2, and $k$). Thus, any even number greater than 2 is a composite number.

Therefore, 2 is the only even number that is prime.

2 is the only prime number which is even.

When we find the factors of two numbers and the only factor they share is 1, these numbers have a special relationship.

Let's consider an example:

Factors of 4 are {1, 2, 4}.

Factors of 9 are {1, 3, 9}.

The common factors of 4 and 9 are the numbers that appear in both lists. The only common factor is 1.

Numbers like 4 and 9, which have only 1 as their common factor, are called coprime numbers or relatively prime numbers.

Two numbers having only 1 as a common factor are called coprime numbers.

To find the number of prime numbers between 1 and 100, we can list the prime numbers in this range.

Remember, a prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's list them:

Primes between 1 and 10: 2, 3, 5, 7 (4 primes)

Primes between 10 and 20: 11, 13, 17, 19 (4 primes)

Primes between 20 and 30: 23, 29 (2 primes)

Primes between 30 and 40: 31, 37 (2 primes)

Primes between 40 and 50: 41, 43, 47 (3 primes)

Primes between 50 and 60: 53, 59 (2 primes)

Primes between 60 and 70: 61, 67 (2 primes)

Primes between 70 and 80: 71, 73, 79 (3 primes)

Primes between 80 and 90: 83, 89 (2 primes)

Primes between 90 and 100: 97 (1 prime)

Total number of primes = $4 + 4 + 2 + 2 + 3 + 2 + 2 + 3 + 2 + 1 = 25$.

The prime numbers between 1 and 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Counting these, there are 25 prime numbers.

Number of primes between 1 to 100 is 25.

A number is divisible by 10 if and only if its last digit is 0.

The last digit of a number is the digit in the ones place.

For example:

The number 50 is divisible by 10 because its ones digit is 0 ($50 \div 10 = 5$).

The number 120 is divisible by 10 because its ones digit is 0 ($120 \div 10 = 12$).

The number 345 is not divisible by 10 because its ones digit is 5.

Thus, for a number to be divisible by 10, the digit in the ones place must be 0.

If a number has 0 in ones place, then it is divisible by 10.

A number is divisible by 5 if and only if its last digit is either 0 or 5.

The last digit of a number is the digit in its ones place.

Examples:

The number 20 is divisible by 5 ($20 \div 5 = 4$). Its ones digit is 0.

The number 35 is divisible by 5 ($35 \div 5 = 7$). Its ones digit is 5.

The number 105 is divisible by 5 ($105 \div 5 = 21$). Its ones digit is 5.

The number 230 is divisible by 5 ($230 \div 5 = 46$). Its ones digit is 0.

The number 123 is not divisible by 5. Its ones digit is 3.

Thus, for a number to be divisible by 5, the digit in its ones place must be either 0 or 5.

A number is divisible by 5, if it has 0 or 5 in its ones place.

A number is divisible by 5 if and only if its last digit is either 0 or 5.

The last digit of a number is the digit in its ones place.

Examples:

The number 20 is divisible by 5 ($20 \div 5 = 4$). Its ones digit is 0.

The number 35 is divisible by 5 ($35 \div 5 = 7$). Its ones digit is 5.

The number 105 is divisible by 5 ($105 \div 5 = 21$). Its ones digit is 5.

The number 230 is divisible by 5 ($230 \div 5 = 46$). Its ones digit is 0.

The number 123 is not divisible by 5. Its ones digit is 3.

Thus, for a number to be divisible by 5, the digit in its ones place must be either 0 or 5.

A number is divisible by 5, if it has 0 or 5 in its ones place.

This is the divisibility rule for 3.

Let's consider some examples:

Consider the number 123.

Sum of digits = $1 + 2 + 3 = 6$.

6 is divisible by 3 ($6 \div 3 = 2$).

Is 123 divisible by 3? Yes, $123 \div 3 = 41$.

Consider the number 549.

Sum of digits = $5 + 4 + 9 = 18$.

18 is divisible by 3 ($18 \div 3 = 6$).

Is 549 divisible by 3? Yes, $549 \div 3 = 183$.

Consider the number 14.

Sum of digits = $1 + 4 = 5$.

5 is not divisible by 3.

Is 14 divisible by 3? No, $14 \div 3 = 4$ with a remainder of 2.

The divisibility rule for 3 states that a number is divisible by 3 if the sum of its digits is divisible by 3.

Another way to say "divisible by 3" is "a multiple of 3".

If the sum of the digits in a number is a multiple of 3, then the number is divisible by 3.

This is the divisibility rule for 11.

The rule states that a number is divisible by 11 if the alternating sum of its digits is divisible by 11. The alternating sum is calculated by taking the difference between the sum of the digits at the odd-numbered places (from the right) and the sum of the digits at the even-numbered places (from the right).

If this difference is 0 or a multiple of 11, then the number is divisible by 11.

Examples:

Consider the number 121.

Digits at odd places (from right): 1 (ones place), 1 (hundreds place). Sum = $1 + 1 = 2$.

Digits at even places (from right): 2 (tens place). Sum = 2.

Difference = Sum of odd place digits - Sum of even place digits = $2 - 2 = 0$.

Since the difference is 0, 121 is divisible by 11 ($121 \div 11 = 11$).

Consider the number 1331.

Digits at odd places (from right): 1 (ones), 3 (hundreds). Sum = $1 + 3 = 4$.

Digits at even places (from right): 3 (tens), 1 (thousands). Sum = $3 + 1 = 4$.

Difference = Sum of odd place digits - Sum of even place digits = $4 - 4 = 0$.

Since the difference is 0, 1331 is divisible by 11 ($1331 \div 11 = 121$).

Consider the number 242.

Digits at odd places (from right): 2 (ones), 2 (hundreds). Sum = $2 + 2 = 4$.

Digits at even places (from right): 4 (tens). Sum = 4.

Difference = $4 - 4 = 0$.

Since the difference is 0, 242 is divisible by 11 ($242 \div 11 = 22$).

Consider the number 616.

Digits at odd places (from right): 6 (ones), 6 (hundreds). Sum = $6 + 6 = 12$.

Digits at even places (from right): 1 (tens). Sum = 1.

Difference = $12 - 1 = 11$.

Since the difference is 11, and 11 is divisible by 11 ($11 \div 11 = 1$), 616 is divisible by 11 ($616 \div 11 = 56$).

Consider the number 10813.

Digits at odd places (from right): 3 (ones), 8 (hundreds), 1 (ten thousands). Sum = $3 + 8 + 1 = 12$.

Digits at even places (from right): 1 (tens), 0 (thousands). Sum = $1 + 0 = 1$.

Difference = $12 - 1 = 11$.

Since the difference is 11, 10813 is divisible by 11 ($10813 \div 11 = 983$).

The condition is that the difference is either 0 or divisible by the number in the blank. From the rule, this number is 11.

If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either 0 or divisible by 11, then the number is divisible by 11.

LCM stands for Least Common Multiple.

The LCM of two or more numbers is defined as the smallest positive integer that is a multiple of all the given numbers.

Let's consider an example with two numbers, say 4 and 6.

Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, ...

Multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, ...

The common multiples of 4 and 6 are the numbers that appear in both lists: 12, 24, 36, ...

The lowest (smallest) of these common multiples is 12.

So, the LCM of 4 and 6 is 12.

The definition explicitly states that the LCM is the lowest of the common "multiples".

The LCM of two or more given numbers is the lowest of their common multiples.

HCF stands for Highest Common Factor.

The HCF of two or more numbers is defined as the largest positive integer that divides each of the given numbers without leaving a remainder.

Let's consider an example with two numbers, say 12 and 18.

Factors of 12 are the numbers that divide 12 exactly: 1, 2, 3, 4, 6, 12.

Factors of 18 are the numbers that divide 18 exactly: 1, 2, 3, 6, 9, 18.

The common factors of 12 and 18 are the numbers that appear in both lists: 1, 2, 3, and 6.

The highest (largest) of these common factors is 6.

So, the HCF of 12 and 18 is 6.

The definition explicitly states that the HCF is the highest of the common "factors".

The HCF of two or more given numbers is the highest of their common factors.

(i) The difference of two consecutive whole numbers: Let the consecutive whole numbers be $n$ and $n+1$. The difference is $(n+1) - n = 1$. This matches with (d) 1.

(ii) The product of two non-zero consecutive whole numbers: Let the numbers be $n$ and $n+1$ where $n \geq 1$. One of these numbers must be even. The product of an even number and any other integer is always even. This matches with (f) even.

(iii) Quotient when zero is divided by another non-zero whole number: Dividing 0 by any non-zero number $a$ gives $\frac{0}{a} = 0$. This matches with (b) 0.

(iv) 2 added three times, to the smallest whole number: The smallest whole number is 0. Adding 2 three times means $2+2+2 = 6$. Adding this to 0 gives $0 + 6 = 6$. This matches with (e) 6.

(v) Smallest odd prime number: Prime numbers are numbers greater than 1 with only factors 1 and themselves. The first few prime numbers are 2, 3, 5, 7, ... The number 2 is even. The smallest prime number that is odd is 3. This matches with (c) 3.

The final matches are:

(i) - (d)

(ii) - (f)

(iii) - (b)

(iv) - (e)

(v) - (c)

To arrange the numbers in descending order, we need to list them from the greatest to the smallest.

The given numbers are: 8435, 4835, 13584, 5348, 25843.

First, let's compare the numbers based on the number of digits they have.

The numbers 8435, 4835, and 5348 are 4-digit numbers.

The numbers 13584 and 25843 are 5-digit numbers.

A number with more digits is generally greater than a number with fewer digits.

Comparing the 5-digit numbers: 13584 and 25843.

Looking at the leftmost digit (Ten Thousands place): 2 in 25843 is greater than 1 in 13584.

So, 25843 is greater than 13584.

Now, comparing the 4-digit numbers: 8435, 4835, and 5348.

Looking at the leftmost digit (Thousands place): Compare 8 (in 8435), 4 (in 4835), and 5 (in 5348).

The greatest digit is 8, followed by 5, then 4.

So, 8435 is the greatest, followed by 5348, and then 4835 among the 4-digit numbers.

Combining the sorted 5-digit and 4-digit numbers in descending order:

The largest number is 25843.

The next largest is 13584.

Then come the 4-digit numbers in descending order: 8435, 5348, and 4835.

Arranging all numbers from greatest to smallest, we get:

25843, 13584, 8435, 5348, 4835.

The given numbers are: 38051425, 30040700, and 67205602.

All three numbers have the same number of digits (8 digits).

To find the greatest and smallest numbers, we compare their digits starting from the leftmost position (the highest place value).

The leftmost digits are:

• For 38051425: 3
• For 30040700: 3
• For 67205602: 6

Comparing these leftmost digits, 6 is the largest digit. So, the number starting with 6 is the greatest.

The greatest number is 67205602.

Now, we compare the other two numbers that start with 3: 38051425 and 30040700.

We look at the next digit to the right (the Ten Lakhs place):

• For 38051425: 8
• For 30040700: 0

Comparing these digits, 0 is smaller than 8. So, the number with 0 in the Ten Lakhs place is the smallest.

The smallest number is 30040700.

The expanded form of a number is written by expressing each digit as the product of the digit and its place value, and then summing these products.

(a) 74836

Place values from right to left are: Ones (1), Tens (10), Hundreds (100), Thousands (1,000), Ten Thousands (10,000).

• Digit 6 is in the Ones place: $6 \times 1$
• Digit 3 is in the Tens place: $3 \times 10$
• Digit 8 is in the Hundreds place: $8 \times 100$
• Digit 4 is in the Thousands place: $4 \times 1000$
• Digit 7 is in the Ten Thousands place: $7 \times 10000$

Expanded form: $7 \times 10000 + 4 \times 1000 + 8 \times 100 + 3 \times 10 + 6 \times 1$

This can also be written as $70000 + 4000 + 800 + 30 + 6$.

(b) 574021

Place values from right to left are: Ones (1), Tens (10), Hundreds (100), Thousands (1,000), Ten Thousands (10,000), Lakhs (100,000).

• Digit 1 is in the Ones place: $1 \times 1$
• Digit 2 is in the Tens place: $2 \times 10$
• Digit 0 is in the Hundreds place: $0 \times 100$
• Digit 4 is in the Thousands place: $4 \times 1000$
• Digit 7 is in the Ten Thousands place: $7 \times 10000$
• Digit 5 is in the Lakhs place: $5 \times 100000$

Expanded form: $5 \times 100000 + 7 \times 10000 + 4 \times 1000 + 0 \times 100 + 2 \times 10 + 1 \times 1$

This can also be written as $500000 + 70000 + 4000 + 0 + 20 + 1$, or simply $500000 + 70000 + 4000 + 20 + 1$.

(c) 8907010

Place values from right to left are: Ones (1), Tens (10), Hundreds (100), Thousands (1,000), Ten Thousands (10,000), Lakhs (100,000), Ten Lakhs (1,000,000).

• Digit 0 is in the Ones place: $0 \times 1$
• Digit 1 is in the Tens place: $1 \times 10$
• Digit 0 is in the Hundreds place: $0 \times 100$
• Digit 7 is in the Thousands place: $7 \times 1000$
• Digit 0 is in the Ten Thousands place: $0 \times 10000$
• Digit 9 is in the Lakhs place: $9 \times 100000$
• Digit 8 is in the Ten Lakhs place: $8 \times 1000000$

Expanded form: $8 \times 1000000 + 9 \times 100000 + 0 \times 10000 + 7 \times 1000 + 0 \times 100 + 1 \times 10 + 0 \times 1$

This can also be written as $8000000 + 900000 + 0 + 7000 + 0 + 10 + 0$, or simply $8000000 + 900000 + 7000 + 10$.

The given populations are:

Maharashtra: 96,878,627

Andhra Pradesh: 76,210,007

Bihar: 82,998,509

Uttar Pradesh: 166,197,921

Let's compare the numbers to arrange them in ascending (smallest to greatest) and descending (greatest to smallest) order.

First, compare the number of digits. Maharashtra, Andhra Pradesh, and Bihar have 8 digits. Uttar Pradesh has 9 digits. A number with more digits is greater.

So, Uttar Pradesh (166,197,921) has the largest population.

Now compare the 8-digit numbers: 96,878,627, 76,210,007, and 82,998,509.

Compare the leftmost digit (Ten Lakhs place): 9 (Maharashtra), 7 (Andhra Pradesh), 8 (Bihar).

The smallest digit is 7, so Andhra Pradesh (76,210,007) has the smallest population among these three.

The next smallest digit is 8, so Bihar (82,998,509) has the next smallest population.

The largest digit among these is 9, so Maharashtra (96,878,627) has the largest population among these three.

Arranging the populations in ascending order:

76,210,007 (Andhra Pradesh)

82,998,509 (Bihar)

96,878,627 (Maharashtra)

166,197,921 (Uttar Pradesh)

Ascending order of states by population:

Andhra Pradesh, Bihar, Maharashtra, Uttar Pradesh.

Arranging the populations in descending order:

166,197,921 (Uttar Pradesh)

96,878,627 (Maharashtra)

82,998,509 (Bihar)

76,210,007 (Andhra Pradesh)

Descending order of states by population:

Uttar Pradesh, Maharashtra, Bihar, Andhra Pradesh.

The given diameter of Jupiter is 142800000 metres.

We need to insert commas according to the International System of Numeration.

In the International System, commas are placed after every three digits from the right, grouping digits into Thousands, Millions, Billions, and so on.

Starting from the rightmost digit and grouping in threes:

142 800 000

Now, insert the commas:

142,800,000

This number is read as one hundred forty-two million, eight hundred thousand.

The diameter according to the International System of Numeration is 142,800,000 metres.

Population in 1961 = 439 millions.

Population in 2001 = 1028 millions.

Total increase in population = Population in 2001 - Population in 1961.

Increase in population = $1028 \text{ millions} - 439 \text{ millions}$.

Let's perform the subtraction:

$\begin{array}{cccc} & 10 & 2 & 8 \\ - & & 4 & 3 & 9 \\ \hline & & 5 & 8 & 9 \\ \hline \end{array}$

Now, we need to write this increase in population in the Indian System of Numeration, using commas suitably.

We know that 1 million = 10 lakh.

So, 589 millions = $589 \times 10$ lakh = 5890 lakh.

We also know that 1 crore = 100 lakh.

5890 lakh can be written as 5890 / 100 crore = 58 crore and 90 lakh.

In terms of the Indian System, 5890 lakh is written as 58,90,00,000 (58 crore 90 lakh).

Alternatively, 1 million is 10 followed by 6 zeros ($1,000,000$).

So, 589 millions = $589 \times 1,000,000 = 589,000,000$.

Now, we write this number in the Indian System. In the Indian System, commas are placed after the hundreds place (3 digits from the right), then after every two digits (Lakhs, Crores, etc.).

Starting from the rightmost digit:

589000000

Group first three digits from the right: 000

Group next two digits: 00

Group next two digits: 90

Remaining digits: 58

Inserting commas: 58,90,00,000.

This is read as fifty-eight crore ninety lakh.

The total increase in population from 1961 to 2001 is 589 million.

The increase in population in Indian System of Numeration is 58,90,00,000.

Given:

Radius of Earth = 6400 km.

Radius of Mars = 4300000 m.

To compare the radii, they must be in the same unit.

Let's convert the radius of Earth from kilometres to metres.

We know that 1 kilometre = 1000 metres.

Now we compare the radii in metres:

Radius of Earth = 6,400,000 metres

Radius of Mars = 4,300,000 metres

Comparing the numbers 6,400,000 and 4,300,000:

Both are 7-digit numbers.

Comparing the leftmost digit (Millions place): 6 (Earth) vs 4 (Mars).

Since 6 is greater than 4, the radius of Earth is bigger than the radius of Mars.

To find by how much the radius of Earth is bigger, we subtract the radius of Mars from the radius of Earth.

Perform the subtraction:

$\begin{array}{ccccccc} & 6 & 4 & 0 & 0 & 0 & 0 & 0 \\ - & 4 & 3 & 0 & 0 & 0 & 0 & 0 \\ \hline & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}$

We can also express this difference in kilometres by dividing by 1000:

The radius of Earth is bigger than the radius of Mars.

The radius of Earth is bigger by 2100000 metres or 2100 km.

The population of Tripura in 2001 was 3,199,203.

This number is given in the International System of Numeration (commas after every three digits).

Reading from left to right, the places are Millions, Hundred Thousands, Ten Thousands, Thousands, Hundreds, Tens, Ones.

3 is in the Millions place.

199 is in the Thousands period (Hundred Thousands, Ten Thousands, Thousands).

203 is in the Units period (Hundreds, Tens, Ones).

So, the number is read as "Three million, one hundred ninety-nine thousand, two hundred three".

Population of Tripura in words: Three million, one hundred ninety-nine thousand, two hundred three.

The population of Meghalaya in 2001 was 2,318,822.

This number is also in the International System of Numeration.

Reading from left to right:

2 is in the Millions place.

318 is in the Thousands period.

822 is in the Units period.

So, the number is read as "Two million, three hundred eighteen thousand, eight hundred twenty-two".

Population of Meghalaya in words: Two million, three hundred eighteen thousand, eight hundred twenty-two.

Number of children who received polio drops in March 2008 = 2,12,583.

Number of children who received polio drops in the next month (April 2008) = 2,16,813.

To find the difference, we subtract the smaller number from the larger number.

Comparing the two numbers: 2,16,813 is greater than 2,12,583.

Perform the subtraction:

$\begin{array}{cccccc} & 2 & 1 & 6 & 8 & 1 & 3 \\ - & 2 & 1 & 2 & 5 & 8 & 3 \\ \hline & & & 4 & 2 & 3 & 0 \\ \hline \end{array}$

Alternatively, using the standard subtraction format:

$\begin{array}{ccccccc} & 2 & 1 & 6 & 8 & 1 & 3 \\ - & 2 & 1 & 2 & 5 & 8 & 3 \\ \hline & 0 & 0 & 4 & 2 & 3 & 0 \\ \hline \end{array}$

The difference is 4230.

The difference of the number of children getting polio drops in the two months is 4230.

Amount of money the person had initially = $\textsf{₹}$ 1,000,000.

Cost of colour T.V. = $\textsf{₹}$ 16,580.

Cost of motor cycle = $\textsf{₹}$ 45,890.

Cost of flat = $\textsf{₹}$ 870,000.

Total amount spent = Cost of T.V. + Cost of motor cycle + Cost of flat.

Let's add the amounts spent:

$\begin{array}{cccccc} & & 1 & 6 & 5 & 8 & 0 \\ & & 4 & 5 & 8 & 9 & 0 \\ + & 8 & 7 & 0 & 0 & 0 & 0 \\ \hline & 9 & 3 & 2 & 4 & 7 & 0 \\ \hline \end{array}$

Amount of money left with him = Initial amount - Total amount spent.

Perform the subtraction:

$\begin{array}{ccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ - & & 9 & 3 & 2 & 4 & 7 & 0 \\ \hline & & 0 & 6 & 7 & 5 & 3 & 0 \\ \hline \end{array}$

Money left with him was $\textsf{₹}$ 67,530.

Total number of Vitamin A tablets = 180000.

Number of tablets distributed = 18734.

To find the number of remaining tablets, we subtract the number of distributed tablets from the total number of tablets.

Perform the subtraction:

$\begin{array}{ccccccc} & 1 & 8 & 0 & 0 & 0 & 0 \\ - & & 1 & 8 & 7 & 3 & 4 \\ \hline & 1 & 6 & 1 & 2 & 6 & 6 \\ \hline \end{array}$

The number of remaining tablets is 161266.

The number of the remaining vitamin tablets is 161266.

Initial amount Chinmay had = $\textsf{₹}$ 610000.

Amount given to Jyoti = $\textsf{₹}$ 87500.

Amount given to Javed = $\textsf{₹}$ 126380.

Amount given to John = $\textsf{₹}$ 350000.

Total amount given away = Amount to Jyoti + Amount to Javed + Amount to John.

Let's add the amounts given away:

$\begin{array}{cccccc} & & 8 & 7 & 5 & 0 & 0 \\ & 1 & 2 & 6 & 3 & 8 & 0 \\ + & 3 & 5 & 0 & 0 & 0 & 0 \\ \hline & 5 & 6 & 3 & 8 & 8 & 0 \\ \hline \end{array}$

Amount of money left with Chinmay = Initial amount - Total amount given away.

Perform the subtraction:

$\begin{array}{ccccccc} & 6 & 1 & 0 & 0 & 0 & 0 \\ - & 5 & 6 & 3 & 8 & 8 & 0 \\ \hline & & 4 & 6 & 1 & 2 & 0 \\ \hline \end{array}$

Money left with him was $\textsf{₹}$ 46,120.

The largest number of seven digits is the number formed by seven 9s.

The smallest number of eight digits is the number formed by a 1 followed by seven 0s.

We need to find the difference between the smallest number of eight digits and the largest number of seven digits.

Perform the subtraction:

$\begin{array}{cccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ - & & 9 & 9 & 9 & 9 & 9 & 9 & 9 \\ \hline & & & & & & & & 1 \\ \hline \end{array}$

The difference is 1.

The difference between the largest number of seven digits and the smallest number of eight digits is 1.

Alternatively, the successor of the largest 7-digit number (9,999,999) is obtained by adding 1, which gives 10,000,000, the smallest 8-digit number. The difference between a number and its predecessor is always 1.

The mobile number has ten digits. Let the number be $d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9 d_{10}$.

The first four digits are given: $d_1 = 9$, $d_2 = 9$, $d_3 = 8$, $d_4 = 7$.

The last three digits are given: $d_8 = 3$, $d_9 = 5$, $d_{10} = 5$.

So the number is $9 \ 9 \ 8 \ 7 \ \text{_____} \ \text{_____} \ \text{_____} \ 3 \ 5 \ 5$.

The digits to be filled in the blanks are $d_5, d_6, d_7$. These are three distinct digits.

We are told that these remaining digits are distinct and make the mobile number the greatest possible number.

To make the entire number as great as possible, we should choose the largest possible digits for the positions $d_5, d_6, d_7$ and place them in descending order.

The digits already used in the number are 9, 9, 8, 7, 3, 5, 5.

The set of digits available for $d_5, d_6, d_7$ are the digits from 0 to 9 that have not been used, considering the requirement that $d_5, d_6, d_7$ must be distinct.

Digits used: {9, 9, 8, 7, 3, 5, 5}

Let's list all possible digits (0 to 9): {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

Let's see which digits from 0 to 9 are available for the positions $d_5, d_6, d_7$. The digits 9, 8, 7, 3, 5 have already appeared at least once.

The available digits are the ones not in the set {9, 8, 7, 3, 5}. These are {0, 1, 2, 4, 6}.

We need to choose three distinct digits from the set {0, 1, 2, 4, 6} to place in positions $d_5, d_6, d_7$ to make the number greatest.

To make the number greatest, the digits in the higher place value positions ($d_5$ first, then $d_6$, then $d_7$) should be as large as possible.

The available distinct digits are 6, 4, 2, 1, 0.

We choose the three largest distinct available digits: 6, 4, and 2.

To make the number greatest, we place these digits in descending order in the positions $d_5, d_6, d_7$.

$d_5 = 6$

$d_6 = 4$

$d_7 = 2$

The mobile number is 9 9 8 7 6 4 2 3 5 5.

The remaining distinct digits are 6, 4, and 2.

These digits are 6, 4, 2.

The mobile number has ten digits: $d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9 d_{10}$.

The first four digits are given: $d_1 = 9$, $d_2 = 9$, $d_3 = 7$, $d_4 = 9$.

So the number starts with 9 9 7 9.

The remaining six digits ($d_5, d_6, d_7, d_8, d_9, d_{10}$) must be chosen from the set {8, 3, 5, 6, 0}.

We are allowed to use exactly one digit from {8, 3, 5, 6, 0} twice. The other four digits must be used once each.

We need to make the smallest possible mobile number.

To make the number smallest, the digits in the positions $d_5, d_6, d_7, d_8, d_9, d_{10}$ should be as small as possible and placed in ascending order.

The available digits to choose from are {8, 3, 5, 6, 0}. We need to use six digits in total, where one digit from this set is repeated once, and the other four are used once.

Let the set of six digits we will use be $S$. The sum of the counts of each digit in $S$ must be 6.

The digits in the set {8, 3, 5, 6, 0} in ascending order are 0, 3, 5, 6, 8.

To make the mobile number smallest (starting from $d_5$), we should use the smallest available digits.

The smallest available digit is 0. If we use 0 twice, the set of digits for $d_5, ..., d_{10}$ would be {0, 0, 3, 5, 6, 8}.

To form the smallest number using these digits, we arrange them in ascending order: 0, 0, 3, 5, 6, 8.

So, $d_5=0, d_6=0, d_7=3, d_8=5, d_9=6, d_{10}=8$.

The resulting mobile number would be 9979003568.

Let's check if using another digit twice results in a smaller number. For example, if we use 3 twice, the digits are {0, 3, 3, 5, 6, 8}. Arranging in ascending order gives 0, 3, 3, 5, 6, 8. The number would be 9979033568, which is larger than 9979003568.

If we use 5 twice, the digits are {0, 3, 5, 5, 6, 8}. Arranging in ascending order gives 0, 3, 5, 5, 6, 8. The number would be 9979035568, also larger.

Using 6 twice: {0, 3, 5, 6, 6, 8}. Number: 9979035668.

Using 8 twice: {0, 3, 5, 6, 8, 8}. Number: 9979035688.

Using 0 twice gives the smallest sequence of digits for the remaining places.

The smallest number will be formed by using the digits {0, 0, 3, 5, 6, 8} in the remaining positions, arranged in ascending order.

The remaining digits are 0, 0, 3, 5, 6, 8. The smallest number is formed by arranging them as 003568.

So the mobile number is 9979003568.

The digits used are 9, 9, 7, 9, 0, 0, 3, 5, 6, 8.

The first four digits are 9, 9, 7, 9. These are given.

The remaining six digits are 0, 0, 3, 5, 6, 8. These are formed by using one digit (0) twice from {8, 3, 5, 6, 0} and the other four (3, 5, 6, 8) once.

The question asks for "What are these digits?". It is asking for the digits used in the remaining six places ($d_5$ to $d_{10}$).

The remaining digits, when making the smallest number, are 0, 0, 3, 5, 6, 8.

The digits used from the set {8, 3, 5, 6, 0} to form these six digits are 0 (twice), 3, 5, 6, 8 (once each).

The question is slightly ambiguous. It could mean the digits used in the remaining places (0, 0, 3, 5, 6, 8) or the specific digits from the set {8, 3, 5, 6, 0} that were used (0, 3, 5, 6, 8). Given the phrasing "make the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0", it seems to be asking for the set of digits used in the remaining six places.

The digits used in the remaining six places are the smallest combination using one digit twice from {8, 3, 5, 6, 0}. This smallest digit is 0. So we use 0 twice, and the rest once: {0, 0, 3, 5, 6, 8}. Arranged in ascending order, these are 0, 0, 3, 5, 6, 8.

These digits are 0, 0, 3, 5, 6, 8 (in order for the smallest number).

Let the five-digit number be represented by its digits in the following places from left to right:

Ten Thousands, Thousands, Hundreds, Tens, Units (Ones).

Let's find the digit for each place based on the given information.

1. "digit at ten’s place is 4":

2. "digit at unit’s place is one fourth of ten’s place digit":

3. "digit at hunderd’s place is 0":

4. "digit at thousand’s place is 5 times of the digit at unit’s place":

5. "ten thousand’s place digit is double the digit at ten’s place":

Now, let's arrange the digits in their respective places:

• Ten Thousands place: 8
• Thousands place: 5
• Hundreds place: 0
• Tens place: 4
• Units place: 1

Putting these digits together, the five-digit number is 85041.

The number is 85041.

Given:

The digits are 2, 0, 4, 7, 6, 5.

Each digit must be used only once to form six-digit numbers.

To Find:

The sum of the greatest and the least six-digit numbers formed by the given digits.

Solution:

To form the greatest six-digit number using the digits 2, 0, 4, 7, 6, 5, we arrange the digits in descending order.

The digits in descending order are 7, 6, 5, 4, 2, 0.

Therefore, the greatest six-digit number formed is 765420.

To form the least six-digit number using the digits 2, 0, 4, 7, 6, 5, we arrange the digits in ascending order, ensuring that the leading digit is not zero.

The digits in ascending order are 0, 2, 4, 5, 6, 7.

Since the number must be a six-digit number, 0 cannot be the first digit. The smallest non-zero digit is 2. We place 2 first, and then arrange the remaining digits (0, 4, 5, 6, 7) in ascending order.

Therefore, the least six-digit number formed is 204567.

Now, we find the sum of the greatest and the least six-digit numbers.

Sum = Greatest number + Least number

Sum = $765420 + 204567$

Sum = $969987$

The sum of the greatest and the least six-digit numbers formed by the digits 2, 0, 4, 7, 6, 5 using each digit only once is 969987.

Given:

Volume of cold drink in the container = 35874 litres.

Capacity of each bottle = 200 ml.

To Find:

The number of bottles of 200 ml capacity that can be filled.

Solution:

First, we need to convert the volume of the cold drink from litres to millilitres, since the bottle capacity is given in millilitres.

We know that 1 litre = 1000 ml.

Total volume of cold drink in ml = $35874 \times 1000$ ml.

Total volume of cold drink = $35874000$ ml.

Now, to find the number of bottles that can be filled, we divide the total volume of cold drink by the capacity of each bottle.

Number of bottles = $\frac{\text{Total volume of cold drink}}{\text{Capacity of each bottle}}$

Number of bottles = $\frac{35874000 \text{ ml}}{200 \text{ ml}}$

We can simplify this by cancelling out common factors.

Number of bottles = $\frac{358740}{2}$

Now, perform the division:

Number of bottles = $179370$

Thus, 179370 bottles of 200 ml capacity each can be filled from the container.

Given:

Total population of the town = 450772.

Ratio of illiterate persons = 1 out of every 14 persons.

To Find:

The total number of illiterate persons in the town.

Solution:

We are given that one out of every 14 persons is illiterate.

This means the fraction of illiterate persons in the town is $\frac{1}{14}$.

To find the number of illiterate persons, we need to calculate $\frac{1}{14}$ of the total population.

Number of illiterate persons = $\frac{1}{14} \times \text{Total population}$

Number of illiterate persons = $\frac{1}{14} \times 450772$

Now, we perform the division:

Number of illiterate persons = $\frac{450772}{14}$

Dividing 450772 by 14:

$\frac{450772}{14} = 32198$

Thus, there are 32198 illiterate persons in the town.

Given:

The numbers are 80, 96, 125, and 160.

To Find:

The Least Common Multiple (LCM) of 80, 96, 125, and 160.

Solution:

We can find the LCM by using the prime factorization method or the division method.

Method 1: Prime Factorization

First, find the prime factorization of each number:

$80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^4 \times 5^1$

$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3^1$

$125 = 5 \times 5 \times 5 = 5^3$

$160 = 2 \times 2 \times 2 \times 2 \times 2 \times 5 = 2^5 \times 5^1$

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

Highest power of 2 is $2^5$

Highest power of 3 is $3^1$

Highest power of 5 is $5^3$

LCM = $2^5 \times 3^1 \times 5^3 = 32 \times 3 \times 125$

LCM = $96 \times 125 = 12000$

Method 2: Division Method

We divide the numbers by common prime factors until all numbers become 1.

LCM = $2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 5 \times 5 = 2^5 \times 3^1 \times 5^3$

LCM = $32 \times 3 \times 125 = 96 \times 125 = 12000$

The LCM of 80, 96, 125, and 160 is 12000.

Given:

We need to form 5-digit numbers using different digits.

The digit 5 must appear at the ten's place.

To Find:

The greatest and the smallest 5-digit numbers formed under the given conditions.

Solution:

A 5-digit number has places: Ten Thousands, Thousands, Hundreds, Tens, Units.

The digit 5 is fixed at the ten's place.

The digits available for the other four places are the remaining nine different digits from 0, 1, 2, 3, 4, 6, 7, 8, 9.

The structure of the number is: $\_$ $\_$ $\_$ 5 $\_$.

Finding the Greatest 5-digit number:

To make the number greatest, we should place the largest available digits in the highest place values (from left to right).

The available digits are 0, 1, 2, 3, 4, 6, 7, 8, 9 (excluding 5).

Ten Thousands place: The largest available digit is 9.

Thousands place: The next largest available digit (excluding 9 and 5) is 8.

Hundreds place: The next largest available digit (excluding 9, 8, and 5) is 7.

Units place: The smallest available digit from the remaining ones (0, 1, 2, 3, 4, 6) should be placed here to maximize the number's value, as the higher place values are already maximized. The smallest remaining digit is 0.

So, the greatest 5-digit number is 98750.

Finding the Smallest 5-digit number:

To make the number smallest, we should place the smallest available digits in the highest place values (from left to right), with the constraint that the leading digit cannot be zero.

The available digits are 0, 1, 2, 3, 4, 6, 7, 8, 9 (excluding 5).

Ten Thousands place: The smallest available non-zero digit is 1.

Thousands place: The smallest available digit from the remaining ones (0, 2, 3, 4, 6, 7, 8, 9, excluding 1 and 5) is 0.

Hundreds place: The next smallest available digit from the remaining ones (2, 3, 4, 6, 7, 8, 9) is 2.

Units place: The next smallest available digit from the remaining ones (3, 4, 6, 7, 8, 9) is 3.

So, the smallest 5-digit number is 10253.

The greatest 5-digit number using different digits with 5 at the ten's place is 98750.

The smallest 5-digit number using different digits with 5 at the ten's place is 10253.

Given:

Initial weight = 2kg 300g

Target weight = 5kg 68g

To Find:

The amount in grams that should be added to 2kg 300g to reach 5kg 68g.

Solution:

First, we need to express both weights in the same unit, grams.

We know that 1 kilogram (kg) = 1000 grams (g).

Convert the initial weight to grams:

$2 \text{ kg } 300 \text{ g} = (2 \times 1000) \text{ g} + 300 \text{ g}$

$= 2000 \text{ g} + 300 \text{ g}$

$= 2300 \text{ g}$.

Convert the target weight to grams:

$5 \text{ kg } 68 \text{ g} = (5 \times 1000) \text{ g} + 68 \text{ g}$

$= 5000 \text{ g} + 68 \text{ g}$

$= 5068 \text{ g}$.

The amount that should be added is the difference between the target weight and the initial weight, both in grams.

Amount to be added = Target weight in grams - Initial weight in grams

Amount to be added = $5068 \text{ g} - 2300 \text{ g}$.

Performing the subtraction:

$5068 - 2300 = 2768$

Amount to be added = $2768$ g.

Therefore, 2768 grams should be added to 2kg 300g to make it 5kg 68g.

Given:

Number of packets in one box = 50

Weight of each biscuit packet = 120 g

Maximum capacity of the van = 900 kg

To Find:

The number of boxes that can be loaded in the van.

Solution:

First, calculate the total weight of one box of biscuits.

Weight of one box = Number of packets per box $\times$ Weight of each packet

Weight of one box = $50 \times 120 \text{ g}$

Weight of one box = $6000 \text{ g}$

Next, convert the weight of one box from grams to kilograms, since the van's capacity is given in kilograms.

We know that 1 kilogram (kg) = 1000 grams (g).

Weight of one box in kg = $\frac{6000 \text{ g}}{1000 \text{ g/kg}}$

Weight of one box = $6 \text{ kg}$

Now, determine how many boxes can be loaded into the van. The van cannot carry beyond 900 kg.

Number of boxes = $\frac{\text{Maximum van capacity}}{\text{Weight of one box}}$

Number of boxes = $\frac{900 \text{ kg}}{6 \text{ kg}}$

Performing the division:

Number of boxes = $150$

Thus, 150 such boxes can be loaded in the van.

Given:

We need to compare the quantities 'lakhs' and 'billions'.

Specifically, we need to find how many lakhs are equivalent to five billions.

To Find:

The number of lakhs that make up five billions.

Solution:

First, let's write down the value of one lakh and one billion in standard form.

In the Indian numbering system:

One lakh = $100,000 = 10^5$.

In the International numbering system:

One billion = $1,000,000,000 = 10^9$.

We want to find how many lakhs are in five billions.

Five billions = $5 \times \text{One billion} = 5 \times 10^9$.

To find how many lakhs make five billions, we divide five billions by the value of one lakh.

Number of lakhs = $\frac{\text{Five billions}}{\text{One lakh}}$

Number of lakhs = $\frac{5 \times 10^9}{10^5}$

Using the property of exponents $\frac{a^m}{a^n} = a^{m-n}$, we get:

Number of lakhs = $5 \times 10^{9-5} = 5 \times 10^4$

$5 \times 10^4 = 5 \times 10000 = 50000$

Therefore, 50000 lakhs make five billions.

Given:

We need to compare the quantities 'millions' and 'crores'.

Specifically, we need to find how many millions are equivalent to 3 crores.

To Find:

The number of millions that make up 3 crores.

Solution:

First, let's write down the value of one crore in the Indian numbering system.

One crore = $1,00,00,000 = 10^7$.

Now, let's write down the value of one million in the International numbering system.

One million = $1,000,000 = 10^6$.

We want to find how many millions are in 3 crores.

Three crores = $3 \times \text{One crore} = 3 \times 10^7$.

To find how many millions make 3 crores, we divide the value of 3 crores by the value of one million.

Number of millions = $\frac{\text{Three crores}}{\text{One million}}$

Number of millions = $\frac{3 \times 10^7}{10^6}$

Using the property of exponents $\frac{a^m}{a^n} = a^{m-n}$, we get:

Number of millions = $3 \times 10^{7-6} = 3 \times 10^1$

$3 \times 10^1 = 3 \times 10 = 30$

Therefore, 30 millions make 3 crores.

To estimate the sum by rounding off each number to the nearest hundreds, we follow the rule:

If the tens digit is 5 or greater, round up the hundreds digit.

If the tens digit is less than 5, keep the hundreds digit the same.

In both cases, replace the tens and units digits with zeros.

(a) 874 + 478

Rounding 874 to the nearest hundred:

The tens digit in 874 is 7, which is greater than or equal to 5. So, we round up the hundreds digit (8) to 9 and replace the tens and units digits with 0s.

874 rounded to the nearest hundred is 900.

Rounding 478 to the nearest hundred:

The tens digit in 478 is 7, which is greater than or equal to 5. So, we round up the hundreds digit (4) to 5 and replace the tens and units digits with 0s.

478 rounded to the nearest hundred is 500.

Estimated sum = $900 + 500 = 1400$.

(b) 793 + 397

Rounding 793 to the nearest hundred:

The tens digit in 793 is 9, which is greater than or equal to 5. So, we round up the hundreds digit (7) to 8 and replace the tens and units digits with 0s.

793 rounded to the nearest hundred is 800.

Rounding 397 to the nearest hundred:

The tens digit in 397 is 9, which is greater than or equal to 5. So, we round up the hundreds digit (3) to 4 and replace the tens and units digits with 0s.

397 rounded to the nearest hundred is 400.

Estimated sum = $800 + 400 = 1200$.

(c) 11244 + 3507

Rounding 11244 to the nearest hundred:

The tens digit in 11244 is 4, which is less than 5. So, we keep the hundreds digit (2) the same and replace the tens and units digits with 0s.

11244 rounded to the nearest hundred is 11200.

Rounding 3507 to the nearest hundred:

The tens digit in 3507 is 0, which is less than 5. So, we keep the hundreds digit (5) the same and replace the tens and units digits with 0s.

3507 rounded to the nearest hundred is 3500.

Estimated sum = $11200 + 3500 = 14700$.

(d) 17677 + 13589

Rounding 17677 to the nearest hundred:

The tens digit in 17677 is 7, which is greater than or equal to 5. So, we round up the hundreds digit (6) to 7 and replace the tens and units digits with 0s.

17677 rounded to the nearest hundred is 17700.

Rounding 13589 to the nearest hundred:

The tens digit in 13589 is 8, which is greater than or equal to 5. So, we round up the hundreds digit (5) to 6 and replace the tens and units digits with 0s.

13589 rounded to the nearest hundred is 13600.

Estimated sum = $17700 + 13600 = 31300$.

To estimate the difference by rounding off each number to the nearest tens, we follow the rule:

If the units digit is 5 or greater, round up the tens digit.

If the units digit is less than 5, keep the tens digit the same.

In both cases, replace the units digit with zero.

(a) 11963 – 9369

Rounding 11963 to the nearest ten:

The units digit in 11963 is 3, which is less than 5. So, we keep the tens digit (6) the same and replace the units digit with 0.

11963 rounded to the nearest ten is 11960.

Rounding 9369 to the nearest ten:

The units digit in 9369 is 9, which is greater than or equal to 5. So, we round up the tens digit (6) to 7 and replace the units digit with 0.

9369 rounded to the nearest ten is 9370.

Estimated difference = $11960 - 9370 = 2590$.

(b) 76877 – 7783

Rounding 76877 to the nearest ten:

The units digit in 76877 is 7, which is greater than or equal to 5. So, we round up the tens digit (7) to 8 and replace the units digit with 0.

76877 rounded to the nearest ten is 76880.

Rounding 7783 to the nearest ten:

The units digit in 7783 is 3, which is less than 5. So, we keep the tens digit (8) the same and replace the units digit with 0.

7783 rounded to the nearest ten is 7780.

Estimated difference = $76880 - 7780 = 69100$.

(c) 10732 – 4354

Rounding 10732 to the nearest ten:

The units digit in 10732 is 2, which is less than 5. So, we keep the tens digit (3) the same and replace the units digit with 0.

10732 rounded to the nearest ten is 10730.

Rounding 4354 to the nearest ten:

The units digit in 4354 is 4, which is less than 5. So, we keep the tens digit (5) the same and replace the units digit with 0.

4354 rounded to the nearest ten is 4350.

Estimated difference = $10730 - 4350 = 6380$.

(d) 78203 – 16407

Rounding 78203 to the nearest ten:

The units digit in 78203 is 3, which is less than 5. So, we keep the tens digit (0) the same and replace the units digit with 0.

78203 rounded to the nearest ten is 78200.

Rounding 16407 to the nearest ten:

The units digit in 16407 is 7, which is greater than or equal to 5. So, we round up the tens digit (0) to 1 and replace the units digit with 0.

16407 rounded to the nearest ten is 16410.

Estimated difference = $78200 - 16410 = 61790$.

To estimate the product by rounding off each number to the nearest tens, we follow the rule:

If the units digit is 5 or greater, round up the tens digit and replace the units digit with 0.

If the units digit is less than 5, keep the tens digit the same and replace the units digit with 0.

(a) 87 × 32

Rounding 87 to the nearest ten:

The units digit is 7 ($ \ge 5 $), so 87 rounds up to 90.

Rounding 32 to the nearest ten:

The units digit is 2 ($ < 5 $), so 32 rounds down to 30.

Estimated product = $90 \times 30$

Estimated product = $2700$.

(b) 311 × 113

Rounding 311 to the nearest ten:

The units digit is 1 ($ < 5 $), so 311 rounds down to 310.

Rounding 113 to the nearest ten:

The units digit is 3 ($ < 5 $), so 113 rounds down to 110.

Estimated product = $310 \times 110$

Estimated product = $34100$.

(c) 3239 × 28

Rounding 3239 to the nearest ten:

The units digit is 9 ($ \ge 5 $), so 3239 rounds up to 3240.

Rounding 28 to the nearest ten:

The units digit is 8 ($ \ge 5 $), so 28 rounds up to 30.

Estimated product = $3240 \times 30$

Estimated product = $97200$.

(d) 1385 × 789

Rounding 1385 to the nearest ten:

The units digit is 5 ($ \ge 5 $), so 1385 rounds up to 1390.

Rounding 789 to the nearest ten:

The units digit is 9 ($ \ge 5 $), so 789 rounds up to 790.

Estimated product = $1390 \times 790$

Estimated product = $1098100$.

Given:

Population in 1991 = 78787.

Population in 2001 = 95833.

Estimation required by rounding off to the nearest hundreds.

To Find:

The estimated increase in population between 1991 and 2001.

Solution:

First, we need to round off the population figures for each year to the nearest hundred.

Rounding rules for nearest hundreds:

Look at the tens digit. If it is 5 or more, round the hundreds digit up and change the tens and units digits to 00. If it is less than 5, keep the hundreds digit the same and change the tens and units digits to 00.

Rounding off the population in 1991 (78787) to the nearest hundred:

The tens digit in 78787 is 8, which is greater than or equal to 5. We round up the hundreds digit (7) to 8 and replace the tens and units digits with 00.

Population in 1991 rounded to the nearest hundred = 78800.

Rounding off the population in 2001 (95833) to the nearest hundred:

The tens digit in 95833 is 3, which is less than 5. We keep the hundreds digit (8) the same and replace the tens and units digits with 00.

Population in 2001 rounded to the nearest hundred = 95800.

Now, we find the estimated increase in population by subtracting the estimated population in 1991 from the estimated population in 2001.

Estimated increase in population = Estimated population in 2001 - Estimated population in 1991

Estimated increase = $95800 - 78800$

Performing the subtraction:

Estimated increase = $17000$

The estimated increase in population between 1991 and 2001, by rounding off each population to the nearest hundreds, is 17000.

Given:

The numbers are 758 and 6784.

To Find:

Estimate the product $758 \times 6784$ using the general rule.

Solution:

The general rule for estimating a product is to round each factor to its greatest place value.

Round 758 to its greatest place value, which is the hundreds place.

The digit in the hundreds place is 7. The digit to its right (tens place) is 5. Since it is 5 or greater, we round up the hundreds digit.

758 rounded to the nearest hundred is 800.

Round 6784 to its greatest place value, which is the thousands place.

The digit in the thousands place is 6. The digit to its right (hundreds place) is 7. Since it is 5 or greater, we round up the thousands digit.

6784 rounded to the nearest thousand is 7000.

Now, estimate the product by multiplying the rounded numbers:

Estimated product = $800 \times 7000$

Estimated product = $8 \times 100 \times 7 \times 1000$

Estimated product = $(8 \times 7) \times (100 \times 1000)$

Estimated product = $56 \times 100000$

Estimated product = $5,600,000$

The estimated product of $758 \times 6784$ using the general rule is 5,600,000.

Given:

Number of shirts produced = 216315.

Number of trousers produced = 182736.

Number of jackets produced = 58704.

To Find:

The total production of all three items in that year.

Solution:

To find the total production, we need to add the number of shirts, trousers, and jackets produced.

Total production = Number of shirts + Number of trousers + Number of jackets

Total production = $216315 + 182736 + 58704$

Adding the numbers:

$216315$

$182736$

$+ \$ 58704$

_________

$457755$

Total production = 457755

The total production of all the three items in that year is 457755.

Given:

The numbers are 160, 170, and 90.

To Find:

The Least Common Multiple (LCM) of 160, 170, and 90.

Solution:

We can find the LCM using the prime factorization method or the division method.

Method 1: Prime Factorization

Find the prime factorization of each number:

$160 = 2 \times 2 \times 2 \times 2 \times 2 \times 5 = 2^5 \times 5^1$

$170 = 2 \times 5 \times 17 = 2^1 \times 5^1 \times 17^1$

$90 = 2 \times 3 \times 3 \times 5 = 2^1 \times 3^2 \times 5^1$

The LCM is the product of the highest powers of all prime factors involved in the factorizations.

Highest power of 2: $2^5$

Highest power of 3: $3^2$

Highest power of 5: $5^1$

Highest power of 17: $17^1$

LCM = $2^5 \times 3^2 \times 5^1 \times 17^1$

LCM = $32 \times 9 \times 5 \times 17$

LCM = $288 \times 85$

LCM = $24480$

Method 2: Division Method

We divide the numbers by common and uncommon prime factors until all numbers are reduced to 1.

LCM = $2 \times 5 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 17$

LCM = $2^5 \times 3^2 \times 5^1 \times 17^1$

LCM = $32 \times 9 \times 5 \times 17 = 24480$

The LCM of 160, 170, and 90 is 24480.

Given:

Total volume of fruit juice in the vessel = 13 litres 200 mL.

Capacity of each glass = 60 mL.

To Find:

The number of glasses of capacity 60 mL that can be filled with the fruit juice.

Solution:

First, we need to convert the total volume of fruit juice into a single unit, millilitres (mL), since the capacity of each glass is given in mL.

We know that 1 litre (L) = 1000 millilitres (mL).

Total volume of fruit juice in mL = 13 litres + 200 mL

$= (13 \times 1000) \text{ mL} + 200 \text{ mL}$

$= 13000 \text{ mL} + 200 \text{ mL}$

$= 13200 \text{ mL}$.

To find the number of glasses that can be filled, we divide the total volume of fruit juice by the capacity of each glass.

Number of glasses = $\frac{\text{Total volume of fruit juice}}{\text{Capacity of each glass}}$

Number of glasses = $\frac{13200 \text{ mL}}{60 \text{ mL}}$

We can simplify this by dividing both numerator and denominator by 10:

Number of glasses = $\frac{1320}{6}$

Now, perform the division:

Number of glasses = $220$

Thus, 220 glasses each of capacity 60 mL can be filled with the fruit juice.

Given:

We need to find the sum of four numbers defined using successor and predecessor operations.

To Find:

The sum of the four numbers.

Solution:

Let's determine each of the four numbers:

(a) Successor of 32:

The successor of a whole number is the number that comes immediately after it. It is obtained by adding 1 to the number.

Successor of 32 = $32 + 1 = 33$.

(b) Predecessor of 49:

The predecessor of a whole number (other than 0) is the number that comes immediately before it. It is obtained by subtracting 1 from the number.

Predecessor of 49 = $49 - 1 = 48$.

(c) Predecessor of the predecessor of 56:

First, find the predecessor of 56.

Predecessor of 56 = $56 - 1 = 55$.

Now, find the predecessor of 55.

Predecessor of 55 = $55 - 1 = 54$.

So, the predecessor of the predecessor of 56 is 54.

(d) Successor of the successor of 67:

First, find the successor of 67.

Successor of 67 = $67 + 1 = 68$.

Now, find the successor of 68.

Successor of 68 = $68 + 1 = 69$.

So, the successor of the successor of 67 is 69.

The four numbers are 33, 48, 54, and 69.

Now, we find the sum of these four numbers:

Sum = $33 + 48 + 54 + 69$

Sum = $81 + 54 + 69$

Sum = $135 + 69$

Sum = $204$

The sum of the four numbers is 204.

Given:

Number of boxes a loading tempo can carry = 482.

Number of boxes a van can carry = 518.

Weight of each box = 15 kg.

To Find:

The total weight that can be carried by both vehicles combined.

Solution:

First, calculate the total weight carried by the loading tempo.

Weight carried by tempo = Number of boxes in tempo $\times$ Weight per box

Weight carried by tempo = $482 \times 15$ kg

$482 \times 15 = 7230$ kg

Next, calculate the total weight carried by the van.

Weight carried by van = Number of boxes in van $\times$ Weight per box

Weight carried by van = $518 \times 15$ kg

$518 \times 15 = 7770$ kg

Finally, find the total weight that can be carried by both vehicles by adding the weight carried by the tempo and the van.

Total weight = Weight carried by tempo + Weight carried by van

Total weight = $7230 \text{ kg} + 7770 \text{ kg}$

$7230 + 7770 = 15000$

Total weight = 15000 kg.

Alternatively, we can find the total number of boxes carried by both vehicles and then multiply by the weight of each box.

Total number of boxes = Number of boxes in tempo + Number of boxes in van

Total number of boxes = $482 + 518 = 1000$ boxes.

Total weight = Total number of boxes $\times$ Weight per box

Total weight = $1000 \times 15 \text{ kg}$

Total weight = $15000 \text{ kg}$.

The total weight that can be carried by both vehicles is 15000 kg.

Given:

Amount spent on food and decoration = $\textsf{₹} \ 216766$

Amount spent on jewellery = $\textsf{₹} \ 122322$

Amount spent on furniture = $\textsf{₹} \ 88234$

Amount spent on kitchen items = $\textsf{₹} \ 26780$

To Find:

The total amount spent by Leela on the above items.

Solution:

To find the total amount spent, we need to add the amounts spent on each item.

Total amount spent = Amount on food and decoration + Amount on jewellery + Amount on furniture + Amount on kitchen items

Total amount spent = $\textsf{₹} \ 216766 + \textsf{₹} \ 122322 + \textsf{₹} \ 88234 + \textsf{₹} \ 26780$

Let's perform the addition:

Total amount spent = $\textsf{₹} \ 454102$.

The total amount spent by Leela on the above items is ₹ 454102.

Given:

Number of strips in one box = 5.

Number of capsules in each strip = 12.

Weight of medicine in each capsule = 500 mg.

Total number of boxes = 32.

To Find:

The total weight in grams of medicine in 32 such boxes.

Solution:

First, calculate the total number of capsules in one box.

Number of capsules in one box = Number of strips per box $\times$ Number of capsules per strip

Number of capsules in one box = $5 \times 12 = 60$ capsules.

Next, calculate the total weight of medicine in one box in milligrams (mg).

Weight of medicine in one box = Number of capsules in one box $\times$ Weight per capsule

Weight of medicine in one box = $60 \times 500 \text{ mg} = 30000 \text{ mg}$.

Now, calculate the total weight of medicine in 32 such boxes in milligrams.

Total weight of medicine in 32 boxes = Number of boxes $\times$ Weight of medicine in one box

Total weight of medicine in 32 boxes = $32 \times 30000 \text{ mg} = 960000 \text{ mg}$.

Finally, convert the total weight from milligrams (mg) to grams (g).

We know that 1 gram (g) = 1000 milligrams (mg).

To convert mg to g, we divide by 1000.

Total weight in grams = $\frac{\text{Total weight in mg}}{1000}$

Total weight in grams = $\frac{960000}{1000} \text{ g}$

Total weight in grams = $960 \text{ g}$.

The total weight in grams of medicine in 32 such boxes is 960 g.

Given:

We are looking for the least number which when divided by 3, 4, and 5 leaves a remainder of 2 in each case.

To Find:

The least such number.

Solution:

Let the required number be $N$.

According to the problem, when $N$ is divided by 3, the remainder is 2. This means $N - 2$ is divisible by 3.

When $N$ is divided by 4, the remainder is 2. This means $N - 2$ is divisible by 4.

When $N$ is divided by 5, the remainder is 2. This means $N - 2$ is divisible by 5.

So, the number $N - 2$ must be a common multiple of 3, 4, and 5.

To find the least such number $N$, the value of $N - 2$ must be the least common multiple (LCM) of 3, 4, and 5.

Let's find the LCM of 3, 4, and 5.

LCM of 3, 4, and 5 = $2 \times 2 \times 3 \times 5 = 4 \times 3 \times 5 = 12 \times 5 = 60$.

The least value of $N - 2$ is 60.

$N - 2 = 60$

$N = 60 + 2$

$N = 62$

Let's verify:

$62 \div 3 = 20$ with remainder $2$ ($3 \times 20 + 2 = 62$).

$62 \div 4 = 15$ with remainder $2$ ($4 \times 15 + 2 = 62$).

$62 \div 5 = 12$ with remainder $2$ ($5 \times 12 + 2 = 62$).

The conditions are satisfied.

The least number which when divided by 3, 4, and 5 leaves remainder 2 in each case is 62.

Given:

Quantity of oil of the first kind = 120 litres.

Quantity of oil of the second kind = 180 litres.

Quantity of oil of the third kind = 240 litres.

The merchant wants to fill the oil in tins of equal capacity.

To Find:

The greatest capacity of such a tin.

Solution:

To find the greatest capacity of a tin that can exactly measure the given quantities of oil (120 L, 180 L, and 240 L), we need to find the greatest common divisor (GCD) or highest common factor (HCF) of these numbers.

The HCF of a set of numbers is the largest number that divides all the numbers in the set without leaving a remainder.

Let's find the prime factorization of each number:

$120 = 2 \times 60 = 2 \times 2 \times 30 = 2 \times 2 \times 2 \times 15 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3^1 \times 5^1$

$180 = 2 \times 90 = 2 \times 2 \times 45 = 2 \times 2 \times 3 \times 15 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5^1$

$240 = 2 \times 120 = 2 \times 2 \times 60 = 2 \times 2 \times 2 \times 30 = 2 \times 2 \times 2 \times 2 \times 15 = 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 2^4 \times 3^1 \times 5^1$

To find the HCF, we take the lowest power of each common prime factor present in all factorizations.

Common prime factors are 2, 3, and 5.

Lowest power of 2 is $2^2$ (from $180$).

Lowest power of 3 is $3^1$ (from $120$ and $240$).

Lowest power of 5 is $5^1$ (from $120$, $180$, and $240$).

HCF = $2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$.

The greatest capacity of the tin should be 60 litres.

This means:

120 litres can be filled using $\frac{120}{60} = 2$ tins.

180 litres can be filled using $\frac{180}{60} = 3$ tins.

240 litres can be filled using $\frac{240}{60} = 4$ tins.

Since 60 divides each quantity exactly, a tin of 60 litres capacity can be used to fill each kind of oil in an exact number of tins, and it is the greatest such capacity.

The greatest capacity of such a tin should be 60 litres.

Given:

We need to form a 4-digit odd number using the digits 1, 2, 4, and 5 only once.

When the first and the last digits of the number are interchanged, the new number is divisible by 4.

To Find:

A 4-digit odd number that satisfies the given conditions.

Solution:

Let the 4-digit number be represented as ABCD, where A, B, C, and D are the digits from the set {1, 2, 4, 5}, used exactly once.

Condition 1: The number must be a 4-digit number. Since none of the digits {1, 2, 4, 5} is 0, any arrangement will be a 4-digit number, provided the first digit is not 0 (which is always true with these digits).

Condition 2: The number must be odd. A number is odd if its units digit (D) is odd. The odd digits in the set {1, 2, 4, 5} are 1 and 5. So, the units digit D must be either 1 or 5.

Condition 3: When the first and last digits are interchanged, the new number is divisible by 4. The new number formed by interchanging the first digit (A) and the last digit (D) will be DBCA.

A number is divisible by 4 if the number formed by its last two digits is divisible by 4. In the new number DBCA, the last two digits form the number CA.

So, the number $10 \times C + A$ must be divisible by 4.

We consider the two possible values for the units digit of the original number (D).

Case 1: The original number ends in 1 (D = 1).

The original number is ABC1. The digits A, B, and C must be a permutation of the remaining digits {2, 4, 5}.

The new number formed by swapping the first and last digits is 1BCA. For this number to be divisible by 4, the number CA must be divisible by 4.

The digits A and C must be from {2, 4, 5}. We need to find pairs (C, A) from {2, 4, 5} such that the 2-digit number CA is divisible by 4.

• If A = 2, C can be 4 or 5.
  • If C = 4, CA = 42. $42 \div 4 = 10$ with remainder 2. Not divisible by 4.
  • If C = 5, CA = 52. $52 \div 4 = 13$. Divisible by 4. Here, A=2, C=5. The remaining digit for B is 4. The original number is 2451. This is odd (ends in 1). The swapped number is 1452, and 52 is divisible by 4. This number satisfies all conditions.
• If A = 4, C can be 2 or 5.
  • If C = 2, CA = 24. $24 \div 4 = 6$. Divisible by 4. Here, A=4, C=2. The remaining digit for B is 5. The original number is 4521. This is odd (ends in 1). The swapped number is 1524, and 24 is divisible by 4. This number satisfies all conditions.
  • If C = 5, CA = 54. $54 \div 4 = 13$ with remainder 2. Not divisible by 4.
• If A = 5, C can be 2 or 4.
  • If C = 2, CA = 25. $25 \div 4 = 6$ with remainder 1. Not divisible by 4.
  • If C = 4, CA = 45. $45 \div 4 = 11$ with remainder 1. Not divisible by 4.

From Case 1, two numbers satisfy the conditions: 2451 and 4521.

Case 2: The original number ends in 5 (D = 5).

The original number is ABC5. The digits A, B, and C must be a permutation of the remaining digits {1, 2, 4}.

The new number formed by swapping the first and last digits is 5BCA. For this number to be divisible by 4, the number CA must be divisible by 4.

The digits A and C must be from {1, 2, 4}. We need to find pairs (C, A) from {1, 2, 4} such that the 2-digit number CA is divisible by 4.

• If A = 1, C can be 2 or 4.
  • If C = 2, CA = 21. Not divisible by 4.
  • If C = 4, CA = 41. Not divisible by 4.
• If A = 2, C can be 1 or 4.
  • If C = 1, CA = 12. $12 \div 4 = 3$. Divisible by 4. Here, A=2, C=1. The remaining digit for B is 4. The original number is 2415. This is odd (ends in 5). The swapped number is 5412, and 12 is divisible by 4. This number satisfies all conditions.
  • If C = 4, CA = 42. Not divisible by 4.
• If A = 4, C can be 1 or 2.
  • If C = 1, CA = 14. Not divisible by 4.
  • If C = 2, CA = 24. $24 \div 4 = 6$. Divisible by 4. Here, A=4, C=2. The remaining digit for B is 1. The original number is 4125. This is odd (ends in 5). The swapped number is 5124, and 24 is divisible by 4. This number satisfies all conditions.

From Case 2, two numbers satisfy the conditions: 2415 and 4125.

We have found four such numbers: 2451, 4521, 2415, and 4125.

The question asks for "a" 4-digit odd number satisfying the conditions. We can provide any one of them.

Let's choose 2451.

Check: The number is 2451. It uses digits 2, 4, 5, 1 once. It is a 4-digit number. It is odd (units digit is 1).

Swap the first and last digits: The new number is 1452. The last two digits are 52. $52 \div 4 = 13$. The new number is divisible by 4.

All conditions are met by the number 2451.

A 4-digit odd number satisfying the conditions is 2451.

Given:

We need to form a 4-digit number using the digits 1, 2, 3, and 4 only once.

The number must be divisible by 4.

To Find:

The smallest such 4-digit number.

Solution:

Let the 4-digit number be represented as ABCD, where A, B, C, and D are the digits 1, 2, 3, and 4 in some order.

A number is divisible by 4 if the number formed by its last two digits (CD) is divisible by 4.

The digits C and D must be distinct digits chosen from the set {1, 2, 3, 4}. The number formed by the last two digits, $10C + D$, must be a multiple of 4.

We need to find the smallest 4-digit number. To do this, we should place the smallest digits in the higher place values (Thousands and Hundreds) as much as possible, while satisfying the divisibility condition for the last two digits.

The digits are {1, 2, 3, 4}. The smallest possible thousands digit (A) is 1.

Let's fix A = 1. The remaining digits for B, C, D are {2, 3, 4}.

The number is 1BCD, where {B, C, D} is a permutation of {2, 3, 4}.

The number CD must be divisible by 4, and C and D must be chosen from {2, 3, 4}.

Let's list the possible 2-digit numbers formed by distinct digits from {2, 3, 4} and check for divisibility by 4:

• Using 2 and 3: 23 (not divisible by 4), 32 ($32 \div 4 = 8$, divisible by 4)
• Using 2 and 4: 24 ($24 \div 4 = 6$, divisible by 4), 42 (not divisible by 4)
• Using 3 and 4: 34 (not divisible by 4), 43 (not divisible by 4)

The possible pairs for the last two digits (C, D) from the set {2, 3, 4} such that CD is divisible by 4 are (3, 2) and (2, 4).

Now we form the 4-digit numbers starting with A=1:

If (C, D) = (3, 2), the digits used are 1, 3, 2. The remaining digit from {2, 3, 4} for the thousands place (B) is 4. The number is 1432.

If (C, D) = (2, 4), the digits used are 1, 2, 4. The remaining digit from {2, 3, 4} for the hundreds place (B) is 3. The number is 1324.

We have two potential numbers that use the digits 1, 2, 3, 4 exactly once and are divisible by 4: 1432 and 1324.

We need the smallest of these two numbers.

Comparing 1432 and 1324, the smallest number is 1324.

Let's check if 1324 is divisible by 4. The last two digits form the number 24. $24 \div 4 = 6$. Yes, 1324 is divisible by 4.

The smallest 4-digit number using each of the digits 1, 2, 3, and 4 only once and divisible by 4 is 1324.

Given:

The postal charges for the three parcels are $\textsf{₹} \ 20$, $\textsf{₹} \ 28$, and $\textsf{₹} \ 36$.

Fatima wants to buy stamps of only one denomination to pay the exact amount for each parcel.

To Find:

The greatest denomination of stamps she must buy.

Solution:

To find the greatest denomination of stamps that can be used to pay exactly for charges of $\textsf{₹} \ 20$, $\textsf{₹} \ 28$, and $\textsf{₹} \ 36$, we need to find the greatest common divisor (GCD) or highest common factor (HCF) of 20, 28, and 36.

The HCF of a set of numbers is the largest number that divides all the numbers in the set without leaving a remainder.

We can find the HCF by listing the factors of each number or by using the prime factorization method.

Method: Prime Factorization

Find the prime factorization of each postal charge amount:

$20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

$28 = 2 \times 2 \times 7 = 2^2 \times 7^1$

$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$

The HCF is the product of the common prime factors raised to the lowest power they appear in any of the factorizations.

The common prime factor is 2. The lowest power of 2 that appears in all factorizations is $2^2$.

The prime factors 3, 5, and 7 are not common to all three numbers.

HCF(20, 28, 36) = $2^2 = 4$.

The greatest denomination of stamps she must buy is $\textsf{₹} \ 4$.

With $\textsf{₹} \ 4$ stamps:

For the first parcel ($\textsf{₹} \ 20$), she needs $\frac{20}{4} = 5$ stamps.

For the second parcel ($\textsf{₹} \ 28$), she needs $\frac{28}{4} = 7$ stamps.

For the third parcel ($\textsf{₹} \ 36$), she needs $\frac{36}{4} = 9$ stamps.

In each case, the payment is exact.

The greatest denomination of stamps she must buy is ₹ 4.

Given:

Brand A biscuits are available in packets of 12.

Brand B biscuits are available in packets of 15.

Brand C biscuits are available in packets of 21.

The shopkeeper wants to buy an equal number of biscuits of each brand.

To Find:

The minimum number of packets of each brand the shopkeeper should buy.

Solution:

The total number of biscuits of each brand must be the same. This total number must be a multiple of the number of biscuits in a packet for each brand (12, 15, and 21).

To find the minimum equal number of biscuits, we need to find the Least Common Multiple (LCM) of 12, 15, and 21.

Let's find the LCM of 12, 15, and 21 using the division method.

LCM(12, 15, 21) = $2 \times 2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 12 \times 35 = 420$.

The minimum equal number of biscuits for each brand is 420.

Now, we calculate the number of packets of each brand required to get 420 biscuits.

Number of packets of Brand A = $\frac{\text{Total biscuits}}{\text{Biscuits per packet of A}} = \frac{420}{12} = 35$ packets.

Number of packets of Brand B = $\frac{\text{Total biscuits}}{\text{Biscuits per packet of B}} = \frac{420}{15} = 28$ packets.

Number of packets of Brand C = $\frac{\text{Total biscuits}}{\text{Biscuits per packet of C}} = \frac{420}{21} = 20$ packets.

The minimum number of packets the shopkeeper should buy is:

Brand A: 35 packets

Brand B: 28 packets

Brand C: 20 packets

Given:

Length of the room floor = 8m 96cm.

Breadth of the room floor = 6m 72cm.

The floor is to be covered by square tiles of the same size.

To Find:

The minimum number of square tiles needed to cover the entire floor.

Solution:

First, convert the dimensions of the room into a single unit, centimeters (cm).

We know that 1 meter (m) = 100 centimeters (cm).

Length = $8 \text{ m } 96 \text{ cm} = (8 \times 100) \text{ cm} + 96 \text{ cm} = 800 \text{ cm} + 96 \text{ cm} = 896 \text{ cm}$.

Breadth = $6 \text{ m } 72 \text{ cm} = (6 \times 100) \text{ cm} + 72 \text{ cm} = 600 \text{ cm} + 72 \text{ cm} = 672 \text{ cm}$.

To cover the rectangular floor exactly with square tiles, the side length of the square tile must be a common divisor of the length and the breadth of the floor.

To minimize the number of tiles, the side length of the square tile must be the greatest possible common divisor of the length and the breadth.

So, the side length of the tile will be the Highest Common Factor (HCF) of 896 cm and 672 cm.

Let's find the HCF of 896 and 672 using the Euclidean algorithm:

$896 = 672 \times 1 + 224$

$672 = 224 \times 3 + 0$

The remainder is 0, so the HCF is the last non-zero remainder, which is 224.

Thus, the greatest possible side length of the square tile is 224 cm.

Now, we calculate the area of the floor and the area of one tile.

Area of the floor = Length $\times$ Breadth = $896 \text{ cm} \times 672 \text{ cm}$.

Area of one tile = Side $\times$ Side = $224 \text{ cm} \times 224 \text{ cm}$.

The minimum number of tiles needed is the ratio of the area of the floor to the area of one tile.

Number of tiles = $\frac{\text{Area of the floor}}{\text{Area of one tile}} = \frac{896 \times 672}{224 \times 224}$

We know that $896 = 4 \times 224$ and $672 = 3 \times 224$.

Number of tiles = $\frac{(4 \times 224) \times (3 \times 224)}{224 \times 224}$

Number of tiles = $\frac{4 \times \cancel{224} \times 3 \times \cancel{224}}{\cancel{224} \times \cancel{224}}$

Number of tiles = $4 \times 3 = 12$.

The minimum number of square tiles of the same size needed to cover the entire floor is 12.

Given:

Number of English books = 780.

Number of Science books = 364.

Each shelf should have the same number of books of each subject.

To Find:

The minimum number of books in each shelf.

Solution:

Let $N$ be the number of shelves.

Let $n_E$ be the number of English books in each shelf.

Let $n_S$ be the number of Science books in each shelf.

According to the problem, each shelf has the same number of English books ($n_E$) and the same number of Science books ($n_S$).

The total number of English books is the number of shelves multiplied by the number of English books per shelf:

The total number of Science books is the number of shelves multiplied by the number of Science books per shelf:

From these equations, it is clear that $N$ must be a common divisor of 780 and 364.

The number of books in each shelf is $n_E + n_S$. We want to find the minimum value of this sum.

From the equations above, $n_E = \frac{780}{N}$ and $n_S = \frac{364}{N}$.

The number of books in each shelf is $n_E + n_S = \frac{780}{N} + \frac{364}{N} = \frac{780 + 364}{N} = \frac{1144}{N}$.

To minimize the value of $\frac{1144}{N}$, the value of $N$ must be as large as possible.

The largest possible value for $N$ is the Highest Common Factor (HCF) of 780 and 364.

Let's find the HCF of 780 and 364 using prime factorization:

Prime factorization of 780:

$\begin{array}{c|cc} 2 & 780 \\ \hline 2 & 390 \\ \hline 3 & 195 \\ \hline 5 & 65 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

$780 = 2 \times 2 \times 3 \times 5 \times 13 = 2^2 \times 3^1 \times 5^1 \times 13^1$

Prime factorization of 364:

$\begin{array}{c|cc} 2 & 364 \\ \hline 2 & 182 \\ \hline 7 & 91 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

$364 = 2 \times 2 \times 7 \times 13 = 2^2 \times 7^1 \times 13^1$

The HCF is the product of the lowest powers of the common prime factors.

Common prime factors are 2 and 13.

Lowest power of 2 is $2^2$.

Lowest power of 13 is $13^1$.

HCF(780, 364) = $2^2 \times 13^1 = 4 \times 13 = 52$.

The maximum number of shelves is $N = 52$.

The minimum number of books in each shelf is $\frac{1144}{N} = \frac{1144}{52}$.

Performing the division:

$\frac{1144}{52} = 22$.

When there are 52 shelves:

Number of English books per shelf $n_E = \frac{780}{52} = 15$.

Number of Science books per shelf $n_S = \frac{364}{52} = 7$.

Total books per shelf = $15 + 7 = 22$.

The minimum number of books in each shelf, such that each shelf has the same number of books of each subject, is 22.

Given:

There are 100 blocks of flats, numbered from 1 to 100.

A school van stops at every sixth block (i.e., block numbers that are multiples of 6).

A school bus stops at every tenth block (i.e., block numbers that are multiples of 10).

Both start from the entrance of the colony (which we can consider as stop 0 or before block 1).

To Find:

The block numbers where both the school van and the school bus will stop.

Solution:

The school van stops at block numbers that are multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, ...

The school bus stops at block numbers that are multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, ...

Both the van and the bus will stop at block numbers that are common multiples of 6 and 10.

To find the common stops, we need to find the common multiples of 6 and 10.

The first common stop (after the start) will be at the Least Common Multiple (LCM) of 6 and 10.

Find the LCM of 6 and 10:

Prime factorization of 6 = $2 \times 3$

Prime factorization of 10 = $2 \times 5$

LCM(6, 10) = $2 \times 3 \times 5 = 30$.

The common stops will occur at multiples of their LCM, which is 30.

The common stops are the multiples of 30 within the range of block numbers from 1 to 100.

The multiples of 30 are: $30 \times 1 = 30$, $30 \times 2 = 60$, $30 \times 3 = 90$, $30 \times 4 = 120$, ...

The block numbers are from 1 to 100. So, the common stops within this range are 30, 60, and 90.

Both the school van and the school bus will stop at blocks numbered 30, 60, and 90.

To test the divisibility of a number by 11, we find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right). If this difference is either 0 or divisible by 11, then the original number is divisible by 11.

(a) 5335

The digits of the number 5335 are 5, 3, 3, 5.

Digits at odd places (1st and 3rd from right): 5 and 3.

Sum of digits at odd places = $5 + 3 = 8$.

Digits at even places (2nd and 4th from right): 3 and 5.

Sum of digits at even places = $3 + 5 = 8$.

Difference = (Sum of digits at odd places) - (Sum of digits at even places)

Difference = $8 - 8 = 0$.

Since the difference is 0, the number 5335 is divisible by 11.

(b) 9020814

The digits of the number 9020814 are 9, 0, 2, 0, 8, 1, 4.

Digits at odd places (1st, 3rd, 5th, 7th from right): 4, 8, 2, 9.

Sum of digits at odd places = $4 + 8 + 2 + 9 = 23$.

Digits at even places (2nd, 4th, 6th from right): 1, 0, 0.

Sum of digits at even places = $1 + 0 + 0 = 1$.

Difference = (Sum of digits at odd places) - (Sum of digits at even places)

Difference = $23 - 1 = 22$.

Since the difference 22 is divisible by 11 ($22 \div 11 = 2$), the number 9020814 is divisible by 11.

Conclusion:

Both 5335 and 9020814 are divisible by 11.

The rule for divisibility by 4 is: A number is divisible by 4 if the number formed by its last two digits is divisible by 4, or if the last two digits are 00.

(a) 4096

The last two digits of the number 4096 form the number 96.

We check if 96 is divisible by 4:

$96 \div 4 = 24$.

Since 96 is divisible by 4, the number 4096 is divisible by 4.

(b) 21084

The last two digits of the number 21084 form the number 84.

We check if 84 is divisible by 4:

$84 \div 4 = 21$.

Since 84 is divisible by 4, the number 21084 is divisible by 4.

(c) 31795012

The last two digits of the number 31795012 form the number 12.

We check if 12 is divisible by 4:

$12 \div 4 = 3$.

Since 12 is divisible by 4, the number 31795012 is divisible by 4.

Conclusion:

All the given numbers 4096, 21084, and 31795012 are divisible by 4.

The rule for divisibility by 9 is: A number is divisible by 9 if the sum of its digits is divisible by 9.

(a) 672

Find the sum of the digits of 672:

Sum of digits = $6 + 7 + 2 = 15$.

Check if the sum of digits (15) is divisible by 9:

$15 \div 9 = 1$ with a remainder of $6$.

Since the sum of the digits (15) is not divisible by 9, the number 672 is not divisible by 9.

(b) 5652

Find the sum of the digits of 5652:

Sum of digits = $5 + 6 + 5 + 2 = 18$.

Check if the sum of digits (18) is divisible by 9:

$18 \div 9 = 2$ with a remainder of $0$.

Since the sum of the digits (18) is divisible by 9, the number 5652 is divisible by 9.

Conclusion:

672 is not divisible by 9.

5652 is divisible by 9.