Chapter 3 Playing With Numbers (Additional Questions)

A factor of a number is a number that divides the given number exactly, leaving no remainder.

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We need to check which of the given options (5, 8, 9, 10) divides 48 without leaving a remainder.

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Let's test each option:

For option (A) 5: $48 \div 5$ gives a quotient of 9 and a remainder of 3. So, 5 is not a factor of 48.

For option (B) 8: $48 \div 8$ gives a quotient of 6 and a remainder of 0. So, 8 is a factor of 48.

For option (C) 9: $48 \div 9$ gives a quotient of 5 and a remainder of 3. So, 9 is not a factor of 48.

For option (D) 10: $48 \div 10$ gives a quotient of 4 and a remainder of 8. So, 10 is not a factor of 48.

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Therefore, the only option that is a factor of 48 is 8.

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The correct option is (B) 8.

A multiple of a number is the product of that number and a positive integer.

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To find the first five multiples of 7, we multiply 7 by the first five positive integers, which are 1, 2, 3, 4, and 5.

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First multiple: $7 \times 1 = 7$

Second multiple: $7 \times 2 = 14$

Third multiple: $7 \times 3 = 21$

Fourth multiple: $7 \times 4 = 28$

Fifth multiple: $7 \times 5 = 35$

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So, the first five multiples of 7 are 7, 14, 21, 28, and 35.

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Comparing this sequence with the given options, we find that option (A) matches the calculated multiples.

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The correct option is (A) 7, 14, 21, 28, 35.

Factors of a number are the numbers that divide it exactly, without leaving a remainder.

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To find the number of factors of 12, we can list all the positive integers that divide 12 evenly:

$12 \div 1 = 12$ (1 and 12 are factors)

$12 \div 2 = 6$ (2 and 6 are factors)

$12 \div 3 = 4$ (3 and 4 are factors)

$12 \div 4 = 3$ (We have already found 4 and 3)

$12 \div 5$ (Leaves a remainder)

$12 \div 6 = 2$ (We have already found 6 and 2)

We only need to check up to the square root of the number, and pair factors. The pairs of factors for 12 are (1, 12), (2, 6), and (3, 4).

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

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Let's count the factors: There are 6 factors.

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Alternate Method (using Prime Factorization):

First, find the prime factorization of 12.

$12 = 2 \times 6 = 2 \times 2 \times 3 = 2^2 \times 3^1$

To find the number of factors, take the exponent of each prime factor, add 1 to each exponent, and multiply the results.

The exponent of 2 is 2. Add 1: $2+1=3$.

The exponent of 3 is 1. Add 1: $1+1=2$.

Number of factors $= (2+1) \times (1+1) = 3 \times 2 = 6$.

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Both methods show that the number 12 has 6 factors.

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The correct option is (C) 6.

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

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Let's examine each option to find its factors:

(A) 9: The factors of 9 are numbers that divide 9 exactly. These are 1, 3, and 9. Since 9 has factors other than 1 and itself (namely 3), it is not a prime number. It is a composite number.

(B) 15: The factors of 15 are 1, 3, 5, and 15. Since 15 has factors other than 1 and itself (namely 3 and 5), it is not a prime number. It is a composite number.

(C) 23: The factors of 23 are numbers that divide 23 exactly. We can check for divisibility by small prime numbers (2, 3, 5, 7, etc.) up to the square root of 23, which is approximately 4.8.

23 is not divisible by 2 (it's odd).

The sum of digits of 23 is $2+3=5$, which is not divisible by 3, so 23 is not divisible by 3.

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

Checking further, the only positive integers that divide 23 exactly are 1 and 23. Thus, 23 has exactly two factors: 1 and itself. Therefore, 23 is a prime number.

(D) 33: The factors of 33 are 1, 3, 11, and 33. Since 33 has factors other than 1 and itself (namely 3 and 11), it is not a prime number. It is a composite number.

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Based on the definition of a prime number and the factors of each option, 23 is the only prime number among the choices.

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The correct option is (C) 23.

A composite number is a natural number greater than 1 that has more than two distinct positive divisors (factors).

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Let's examine the first few natural numbers greater than 1:

The number 1: It has only one factor, which is 1. By definition, it is neither prime nor composite.

The number 2: Its factors are 1 and 2. It has exactly two distinct positive divisors, so it is a prime number.

The number 3: Its factors are 1 and 3. It has exactly two distinct positive divisors, so it is a prime number.

The number 4: Its factors are 1, 2, and 4. It has more than two positive divisors (1, 2, and 4), so it is a composite number.

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The smallest natural number greater than 1 is 2. 2 is prime. The next is 3, which is also prime. The next is 4, which is composite.

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Therefore, the smallest composite number is 4.

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The correct option is (D) 4.

A pair of twin primes is a pair of prime numbers that differ by 2.

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We need to check each given pair to see if both numbers are prime and if their difference is 2.

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(A) (2, 3):

2 is a prime number (factors are 1 and 2).

3 is a prime number (factors are 1 and 3).

The difference between the numbers is $3 - 2 = 1$. Since the difference is not 2, this is not a pair of twin primes.

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(B) (3, 5):

3 is a prime number (factors are 1 and 3).

5 is a prime number (factors are 1 and 5).

The difference between the numbers is $5 - 3 = 2$. Since both numbers are prime and their difference is 2, this is a pair of twin primes.

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(C) (4, 5):

4 is not a prime number (factors are 1, 2, and 4).

5 is a prime number (factors are 1 and 5).

Since 4 is not prime, this pair cannot be a pair of twin primes.

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(D) (5, 11):

5 is a prime number (factors are 1 and 5).

11 is a prime number (factors are 1 and 11).

The difference between the numbers is $11 - 5 = 6$. Since the difference is not 2, this is not a pair of twin primes.

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Based on the analysis, only the pair (3, 5) satisfies the condition for being twin primes.

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The correct option is (B) (3, 5).

The divisibility rule for a composite number states that a number is divisible by a composite number if and only if it is divisible by its coprime factors.

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The number 6 can be factored into its prime factors: $6 = 2 \times 3$. The prime factors 2 and 3 are coprime because their greatest common divisor is 1.

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Therefore, a number is divisible by 6 if and only if it is divisible by both 2 and 3.

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Let's examine the options:

(A) 2 and 3: If a number is divisible by both 2 and 3, it is divisible by their product $2 \times 3 = 6$. This is the correct divisibility rule for 6.

(B) 2 and 4: If a number is divisible by 4, it is automatically divisible by 2. Checking divisibility by 2 and 4 is redundant. Moreover, being divisible by 2 and 4 does not guarantee divisibility by 6 (e.g., 4 is divisible by 2 and 4, but not by 6).

(C) 3 and 4: If a number is divisible by both 3 and 4, it is divisible by their least common multiple, which is $\text{LCM}(3, 4) = 12$. A number divisible by 12 is also divisible by 6, but the rule for 6 is specifically about divisibility by 2 and 3. For example, 9 is divisible by 3 but not by 4, and thus not by 6. 10 is divisible by 2 but not by 3, and thus not by 6. This condition (divisibility by 3 and 4) is equivalent to divisibility by 12.

(D) 2, 3 and 6: If a number is divisible by 6, it is necessarily divisible by its factors 2 and 3. So, checking divisibility by 2 and 3 is sufficient. Checking divisibility by 6 is what we are trying to determine, so including 6 in the condition makes it circular. The core rule is based on the prime factors 2 and 3.

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The standard divisibility rule for 6 states that a number must be divisible by both 2 and 3.

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The correct option is (A) 2 and 3.

A number is divisible by 9 if the sum of its digits is divisible by 9.

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Let's find the sum of the digits for each option:

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(A) 123: Sum of digits $= 1 + 2 + 3 = 6$.

6 is not divisible by 9. So, 123 is not divisible by 9.

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(B) 234: Sum of digits $= 2 + 3 + 4 = 9$.

9 is divisible by 9 ($9 \div 9 = 1$). So, 234 is divisible by 9.

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(C) 345: Sum of digits $= 3 + 4 + 5 = 12$.

12 is not divisible by 9. So, 345 is not divisible by 9.

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(D) 456: Sum of digits $= 4 + 5 + 6 = 15$.

15 is not divisible by 9. So, 456 is not divisible by 9.

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Only the number 234 has a sum of digits that is divisible by 9.

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The correct option is (B) 234.

We are given the divisibility rule for 11: 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.

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Let's apply this rule to each given number:

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(A) 1331:

Sum of digits at odd places (1st and 3rd from right): $1 + 3 = 4$

Sum of digits at even places (2nd and 4th from right): $3 + 1 = 4$

Difference = $4 - 4 = 0$

Since the difference is 0, the number 1331 is divisible by 11.

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(B) 1234:

Sum of digits at odd places (1st and 3rd from right): $4 + 2 = 6$

Sum of digits at even places (2nd and 4th from right): $3 + 1 = 4$

Difference = $6 - 4 = 2$

Since the difference is 2 (which is neither 0 nor a multiple of 11), the number 1234 is not divisible by 11.

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(C) 1452:

Sum of digits at odd places (1st and 3rd from right): $2 + 4 = 6$

Sum of digits at even places (2nd and 4th from right): $5 + 1 = 6$

Difference = $6 - 6 = 0$

Since the difference is 0, the number 1452 is divisible by 11.

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(D) 1563:

Sum of digits at odd places (1st and 3rd from right): $3 + 5 = 8$

Sum of digits at even places (2nd and 4th from right): $6 + 1 = 7$

Difference = $8 - 7 = 1$

Since the difference is 1 (which is neither 0 nor a multiple of 11), the number 1563 is not divisible by 11.

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Based on the divisibility rule, both 1331 and 1452 are divisible by 11. In a standard multiple-choice question, there is usually only one correct answer. However, based strictly on the options and the rule provided, both (A) and (C) are divisible by 11.

Assuming this is a question with a single intended answer, and both options satisfy the condition, there might be an error in the question or options. If we must select one, we will choose option (A) as it is a well-known example ($1331 = 11^3$).

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The correct option is (A) 1331.

We are asked to identify the incorrect divisibility test among the given options.

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Let's evaluate each statement:

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(A) A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.

This statement describes the correct divisibility rule for 2. Numbers ending in an even digit are divisible by 2.

This statement is correct.

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(B) A number is divisible by 3 if the sum of its digits is divisible by 3.

This statement describes the correct divisibility rule for 3. For example, for the number 123, the sum of digits is $1+2+3=6$, which is divisible by 3, and 123 is divisible by 3 ($123 \div 3 = 41$).

This statement is correct.

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(C) A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

This statement describes the correct divisibility rule for 4. For example, for the number 1348, the number formed by the last two digits is 48. 48 is divisible by 4 ($48 \div 4 = 12$), and 1348 is divisible by 4 ($1348 \div 4 = 337$).

This statement is correct.

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(D) A number is divisible by 5 if its last digit is 0 or 1.

The correct divisibility rule for 5 states that a number is divisible by 5 if its last digit is 0 or 5. For example, 10 is divisible by 5, and 15 is divisible by 5. However, a number ending in 1, like 11 or 21, is not divisible by 5.

This statement is incorrect.

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Comparing all the statements, option (D) presents an incorrect divisibility rule for 5.

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The correct option is (D) A number is divisible by 5 if its last digit is 0 or 1.

Prime factorization is the process of finding the prime numbers that multiply together to make the original number.

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We need to express 36 as a product of prime numbers.

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Let's find the prime factors of 36:

Start dividing 36 by the smallest prime number, 2:

$36 \div 2 = 18$

Divide 18 by 2:

$18 \div 2 = 9$

9 is not divisible by 2, so move to the next prime number, 3:

$9 \div 3 = 3$

3 is a prime number.

So, the prime factors of 36 are 2, 2, 3, and 3.

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Writing 36 as a product of its prime factors:

$36 = 2 \times 2 \times 3 \times 3$

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Using exponents to represent the repeated factors:

$36 = 2^2 \times 3^2$

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Now let's examine the given options:

(A) $2 \times 18$: 18 is not a prime number ($18 = 2 \times 3^2$). This is not the prime factorization.

(B) $3 \times 12$: 12 is not a prime number ($12 = 2^2 \times 3$). This is not the prime factorization.

(C) $2^2 \times 3^2$: This is $(2 \times 2) \times (3 \times 3) = 4 \times 9 = 36$. The factors 2 and 3 are prime. This is the prime factorization of 36.

(D) $4 \times 9$: Neither 4 nor 9 is a prime number ($4=2^2$, $9=3^2$). This is not the prime factorization.

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The prime factorization of 36 is $2^2 \times 3^2$.

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The correct option is (C) $2^2 \times 3^2$.

The Highest Common Factor (HCF) of two or more numbers is the largest number that divides each of the numbers exactly.

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Method 1: Listing Factors

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

Factors of 18 are: 1, 2, 3, 6, 9, 18.

The common factors of 12 and 18 are 1, 2, 3, and 6.

The highest among the common factors is 6.

So, HCF(12, 18) = 6.

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Method 2: Prime Factorization

First, find the prime factorization of each number.

Prime factorization of 12:

$12 = 2 \times 6 = 2 \times 2 \times 3 = 2^2 \times 3^1$

Prime factorization of 18:

$18 = 2 \times 9 = 2 \times 3 \times 3 = 2^1 \times 3^2$

To find the HCF, take the lowest power of each common prime factor and multiply them.

Common prime factors are 2 and 3.

Lowest power of 2 is $2^1$.

Lowest power of 3 is $3^1$.

HCF$(12, 18) = 2^1 \times 3^1 = 2 \times 3 = 6$.

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Both methods give the same result: the HCF of 12 and 18 is 6.

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The correct option is (B) 6.

The Least Common Multiple (LCM) of two or more numbers is the smallest positive integer that is a multiple of all the numbers.

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Method 1: Listing Multiples

List the multiples of 8:

$8 \times 1 = 8$

$8 \times 2 = 16$

$8 \times 3 = 24$

$8 \times 4 = 32$

$8 \times 5 = 40$

Multiples of 8: 8, 16, 24, 32, 40, ...

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List the multiples of 12:

$12 \times 1 = 12$

$12 \times 2 = 24$

$12 \times 3 = 36$

$12 \times 4 = 48$

$12 \times 5 = 60$

Multiples of 12: 12, 24, 36, 48, 60, ...

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The common multiples are the numbers that appear in both lists. The smallest common multiple is the LCM.

The common multiples are 24, 48, ...

The smallest common multiple is 24.

So, LCM(8, 12) = 24.

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Method 2: Prime Factorization

Find the prime factorization of each number.

Prime factorization of 8:

$8 = 2 \times 2 \times 2 = 2^3$

Prime factorization of 12:

$12 = 2 \times 2 \times 3 = 2^2 \times 3^1$

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To find the LCM, take the highest power of all prime factors that appear in the factorization of either number and multiply them together.

The prime factors involved are 2 and 3.

The highest power of 2 is $2^3$ (from the factorization of 8).

The highest power of 3 is $3^1$ (from the factorization of 12).

LCM$(8, 12) = 2^3 \times 3^1 = 8 \times 3 = 24$.

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Method 3: Using the formula LCM$(a, b) = \frac{a \times b}{\text{HCF}(a, b)}$

First, find the HCF of 8 and 12.

Factors of 8: 1, 2, 4, 8

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

The common factors are 1, 2, 4. The highest common factor is 4.

HCF$(8, 12) = 4$.

Now, use the formula:

LCM$(8, 12) = \frac{8 \times 12}{4} = \frac{96}{4} = 24$.

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All methods show that the LCM of 8 and 12 is 24.

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The correct option is (B) 24.

Given:

HCF of two numbers = 6

LCM of two numbers = 36

One number = 12

To Find: The other number.

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Solution:

We know the relationship between HCF, LCM, and two positive numbers. For any two positive integers $a$ and $b$, their product is equal to the product of their HCF and LCM.

$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$

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Let the two numbers be $a$ and $b$. We are given $a = 12$, $\text{HCF}(a, b) = 6$, and $\text{LCM}(a, b) = 36$. We need to find $b$.

Substitute the given values into the formula:

$12 \times b = 6 \times 36$

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Now, solve for $b$:

$b = \frac{6 \times 36}{12}$

Simplify the expression:

$b = \frac{6}{\cancel{12}_{2}} \times \cancel{36}^{3}$

$b = 6 \times 3 \div 1$ (Mistake in cancellation representation, let's redo the calculation clearly)

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Let's perform the calculation:

$b = \frac{6 \times 36}{12}$

$b = \frac{216}{12}$

Performing the division:

$216 \div 12$

$216 \div 12 = 18$

Alternatively, simplify $\frac{36}{12} = 3$, then $b = 6 \times 3 = 18$.

The other number is 18.

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The correct option is (A) 18.

We are given an Assertion (A) and a Reason (R) and need to determine their truthfulness and whether R correctly explains A.

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Assertion (A): 2 is the only even prime number.

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.

An even number is an integer that is divisible by 2.

The number 2 is greater than 1. Its positive divisors are 1 and 2. Since it has exactly two distinct positive divisors, 2 is a prime number.

The number 2 is also divisible by 2, so it is an even number.

Consider any other even number, say $n$, where $n > 2$. Since $n$ is even, it is divisible by 2. This means that 2 is a positive divisor of $n$. Since $n > 2$, 2 is a positive divisor of $n$ that is different from 1 and $n$. Thus, any even number $n > 2$ has at least three distinct positive divisors: 1, 2, and $n$. By the definition of a prime number, a number with more than two positive divisors is a composite number.

Therefore, no even number greater than 2 can be a prime number.

Thus, 2 is indeed the only even prime number.

Assertion (A) is True.

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Reason (R): All other even numbers are divisible by 2 and themselves.

"Other even numbers" refers to even numbers excluding 2 (i.e., 4, 6, 8, 10, ...).

Any even number is, by definition, divisible by 2.

Any number $n$ is divisible by itself (as $n \div n = 1$).

So, for any even number $n$, it is divisible by 2 and by itself ($n$). The statement is true for all even numbers, including those other than 2.

Reason (R) is True.

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Now, let's check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that 2 is the *only* even prime number. Reason (R) states that all other even numbers are divisible by 2 and themselves. The fact that other even numbers are divisible by 2 (in addition to 1 and themselves) is precisely why they are not prime (provided they are greater than 2). Having 2 as a factor, besides 1 and the number itself, means they have more than two factors, making them composite.

Therefore, Reason (R) correctly explains why even numbers greater than 2 are not prime, and thus why 2 is the only even prime number.

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Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

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The correct option is (A) Both A and R are true and R is the correct explanation of A.

A number is divisible by 3 if the sum of its digits is divisible by 3.

A number is divisible by 5 if its last digit is 0 or 5.

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For a number to be divisible by both 3 and 5, it must satisfy both divisibility rules. Alternatively, since 3 and 5 are coprime, a number divisible by both 3 and 5 must be divisible by their product, $3 \times 5 = 15$. We can check which of the given numbers are divisible by 15.

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Let's check each option:

(A) 30:

Sum of digits $= 3 + 0 = 3$. 3 is divisible by 3.

Last digit is 0. 30 is divisible by 5.

Since 30 is divisible by both 3 and 5, it is divisible by 15 ($30 \div 15 = 2$).

30 is divisible by both 3 and 5.

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(B) 45:

Sum of digits $= 4 + 5 = 9$. 9 is divisible by 3.

Last digit is 5. 45 is divisible by 5.

Since 45 is divisible by both 3 and 5, it is divisible by 15 ($45 \div 15 = 3$).

45 is divisible by both 3 and 5.

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(C) 55:

Sum of digits $= 5 + 5 = 10$. 10 is not divisible by 3.

Last digit is 5. 55 is divisible by 5.

Since 55 is not divisible by 3, it is not divisible by both 3 and 5 (and not divisible by 15).

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(D) 75:

Sum of digits $= 7 + 5 = 12$. 12 is divisible by 3.

Last digit is 5. 75 is divisible by 5.

Since 75 is divisible by both 3 and 5, it is divisible by 15 ($75 \div 15 = 5$).

75 is divisible by both 3 and 5.

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The numbers that are divisible by both 3 and 5 are 30, 45, and 75.

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The correct options are (A) 30, (B) 45, and (D) 75.

We are looking for the smallest number that is a multiple of both 4 and 6. This number is known as the Least Common Multiple (LCM) of 4 and 6.

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To find the LCM, we can list the multiples of each number until we find the first common one.

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Multiples of 4: 4, 8, 12, 16, 20, 24, ...

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Multiples of 6: 6, 12, 18, 24, 30, ...

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The common multiples are the numbers that appear in both lists. The smallest number that appears in both lists is 12.

Alternatively, using prime factorization:

$4 = 2 \times 2 = 2^2$

$6 = 2 \times 3 = 2^1 \times 3^1$

The LCM is found by taking the highest power of all prime factors present in either number.

Highest power of 2 is $2^2 = 4$.

Highest power of 3 is $3^1 = 3$.

LCM$(4, 6) = 2^2 \times 3^1 = 4 \times 3 = 12$.

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Both methods show that the smallest number that is a multiple of both 4 and 6 is 12.

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The correct option is (B) 12.

Let's determine the properties of each number provided:

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(i) 13:

Factors of 13 are 1 and 13. Since it has exactly two distinct positive divisors (1 and itself) and is greater than 1, 13 is a prime number.

So, (i) matches (c).

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(ii) 15:

Factors of 15 are 1, 3, 5, and 15. Since it has more than two distinct positive divisors and is greater than 1, 15 is a composite number.

The last digit of 15 is 5. A number is divisible by 5 if its last digit is 0 or 5. Therefore, 15 is a multiple of 5.

So, (ii) matches (a) and (d).

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(iii) 1:

The number 1 has only one positive divisor, which is 1. By definition, 1 is neither prime nor composite.

So, (iii) matches (b).

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(iv) 4:

Factors of 4 are 1, 2, and 4. Since it has more than two distinct positive divisors and is greater than 1, 4 is a composite number.

So, (iv) matches (a).

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Based on our analysis:

(i) 13 is a Prime number (c).

(iii) 1 is Neither prime nor composite (b).

(iv) 4 is a Composite number (a).

This leaves (ii) 15 to be matched with the remaining property (d) Multiple of 5. As we determined, 15 is indeed a multiple of 5.

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The pairings that uniquely match each number to a property from the list are:

(i) - (c)

(ii) - (d)

(iii) - (b)

(iv) - (a)

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Let's check which option contains these pairings:

(A) (i)-(c), (ii)-(a), (iii)-(b), (iv)-(a)

(B) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)

(C) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)

(D) (i)-(a), (ii)-(c), (iii)-(d), (iv)-(b)

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Option (B) matches the pairings we derived: (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a).

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The correct option is (B) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a).

The teacher needs to distribute 48 pencils and 36 erasers equally among the maximum possible number of students without any left over.

This means the number of students must be a common divisor of both 48 and 36.

To find the maximum number of students, we need to find the Highest Common Factor (HCF) of 48 and 36.

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Let's find the HCF of 48 and 36 using the prime factorization method.

Prime factorization of 48:

$\begin{array}{c|cc} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3^1$.

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Prime factorization of 36:

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

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

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To find the HCF, we take the lowest power of each common prime factor.

The common prime factors are 2 and 3.

Lowest power of 2 is $2^2$.

Lowest power of 3 is $3^1$.

HCF$(48, 36) = 2^2 \times 3^1 = 4 \times 3 = 12$.

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The HCF of 48 and 36 is 12.

Therefore, the maximum number of students among whom the items can be distributed equally is 12.

With 12 students, each student would receive $48 \div 12 = 4$ pencils and $36 \div 12 = 3$ erasers.

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The correct option is (B) 12.

A factor of a number is a number that divides the given number exactly, leaving no remainder.

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Let's consider each option:

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(A) 0: Division by zero is undefined. 0 cannot be a factor of any non-zero number.

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(B) 1: For any integer $n$, $n \div 1 = n$. The remainder is always 0. This means that 1 divides every integer exactly. Therefore, 1 is a factor of every number.

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(C) The number itself: A number is always a factor of itself (since $n \div n = 1$). However, it is not a factor of *every* other number. For example, 5 is a factor of 5, but it is not a factor of 7.

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(D) 2: 2 is a factor of even numbers, but it is not a factor of odd numbers (e.g., 3, 5, 7, etc.). So, 2 is not a factor of every number.

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Based on the definition of a factor, the only number among the options that divides every number exactly is 1.

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The correct option is (B) 1.

A multiple of a number $n$ is any number that can be obtained by multiplying $n$ by an integer $k$. In other words, a number $m$ is a multiple of $n$ if $m = n \times k$ for some integer $k$.

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We need to find which of the given options is a multiple of every number.

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Let's consider each option:

(A) 0: Is 0 a multiple of every number? Let $n$ be any number. We need to check if $0 = n \times k$ for some integer $k$.

If $n \neq 0$, we can choose $k=0$. Then $n \times 0 = 0$. So, 0 is a multiple of any non-zero number.

If $n = 0$, the equation becomes $0 = 0 \times k$. This is true for any integer $k$. So, 0 is also a multiple of 0.

Thus, 0 is a multiple of every number.

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(B) 1: Is 1 a multiple of every number? Let $n$ be any number. We need to check if $1 = n \times k$ for some integer $k$.

If $n=1$, $1 = 1 \times 1$, so $k=1$. Yes.

If $n=2$, $1 = 2 \times k$. For $k$ to be an integer, this is not possible. $k = \frac{1}{2}$ is not an integer.

Therefore, 1 is not a multiple of every number (e.g., it's not a multiple of 2, 3, etc., in the context of integers).

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(C) The number itself: This option is phrased generally. It suggests that if we take a number, say $m$, then $m$ is a multiple of every number $n$. This means $m = n \times k$ for all possible values of $n$. This is not true. For example, 5 is not a multiple of 2.

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(D) Itself and 1: This option refers to a set of numbers ("Itself and 1") rather than a single number that is a multiple of every other number. It does not fit the structure of the question asking for "Which of the following numbers".

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Based on the analysis, only 0 fits the description of being a multiple of every number.

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The correct option is (A) 0.

Co-prime numbers (or relatively prime numbers) are two integers that have no common positive divisors other than 1.

The Highest Common Factor (HCF) of two or more numbers is the largest positive integer that divides each of the numbers exactly.

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By the definition of co-prime numbers, their only positive common divisor is 1.

Since 1 is the only positive common divisor, it is also the largest positive common divisor.

Therefore, the HCF of two co-prime numbers is always 1.

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Example:

Consider the numbers 7 and 10. These are co-prime because their only common positive divisor is 1.

Factors of 7: 1, 7

Factors of 10: 1, 2, 5, 10

The common factors are just {1}. The highest common factor is 1.

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Let's examine the options based on the definition:

(A) 0: HCF is always a positive integer for non-zero numbers. 0 cannot be the HCF.

(B) 1: This matches our conclusion based on the definition of co-prime numbers.

(C) The product of the numbers: The product of two co-prime numbers is equal to their LCM, not their HCF (unless one of the numbers is 1, in which case the HCF is 1 and the product is the other number, which might be the LCM). For example, HCF(7, 10) = 1, Product = $7 \times 10 = 70$. $1 \neq 70$.

(D) The sum of the numbers: The sum of two co-prime numbers is not generally their HCF. For example, HCF(7, 10) = 1, Sum = $7 + 10 = 17$. $1 \neq 17$.

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The HCF of two co-prime numbers is definitively 1.

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The correct option is (B) 1.

Co-prime numbers are two integers that have no common positive divisors other than 1. Their Highest Common Factor (HCF) is 1.

The Least Common Multiple (LCM) of two numbers is the smallest positive integer that is a multiple of both numbers.

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There is a fundamental relationship between the HCF and LCM of two positive integers $a$ and $b$. The product of the two numbers is equal to the product of their HCF and LCM.

$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$

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Let the two co-prime numbers be $a$ and $b$. By the definition of co-prime numbers, their HCF is 1.

So, $\text{HCF}(a, b) = 1$.

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Substitute the HCF value into the formula relating HCF, LCM, and the product of the numbers:

$a \times b = 1 \times \text{LCM}(a, b)$

$a \times b = \text{LCM}(a, b)$

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This shows that the LCM of two co-prime numbers is equal to their product.

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Example:

Consider the co-prime numbers 4 and 9.

HCF(4, 9) = 1.

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, ...

Multiples of 9: 9, 18, 27, 36, 45, ...

The smallest common multiple is 36. So, LCM(4, 9) = 36.

The product of the numbers is $4 \times 9 = 36$.

In this example, LCM(4, 9) = $4 \times 9 = 36$, which confirms the relationship.

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Based on the relationship between HCF and LCM, the LCM of two co-prime numbers is equal to their product.

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The correct option is (C) The product of the numbers.

To determine the approximate value of the square root of 101, we can calculate the squares of the given options and see which is closest to 101.

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Let's square each option:

(A) 9: $9^2 = 9 \times 9 = 81$

(B) 10: $10^2 = 10 \times 10 = 100$

(C) 11: $11^2 = 11 \times 11 = 121$

(D) 12: $12^2 = 12 \times 12 = 144$

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Now, let's compare these squared values to 101:

The difference between 101 and 81 is $101 - 81 = 20$.

The difference between 101 and 100 is $101 - 100 = 1$.

The difference between 121 and 101 is $121 - 101 = 20$.

The difference between 144 and 101 is $144 - 101 = 43$.

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The square of 10 (which is 100) is the closest to 101 among the squares of the given options, with a difference of only 1.

Thus, the approximate value of $\sqrt{101}$ is closest to 10.

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The correct option is (B) 10.

We need to identify the correct divisibility rule for the number 8.

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Let's consider the divisibility rule for 8:

A positive integer is divisible by 8 if and only if the number formed by its last three digits is divisible by 8.

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Let's examine the given options:

(A) The sum of its digits is divisible by 8. This is the divisibility rule for 9 (or 3). It is not the rule for 8. For example, the sum of digits of 16 is $1+6=7$, which is not divisible by 8, but 16 is divisible by 8. The sum of digits of 8 is 8, which is divisible by 8, and 8 is divisible by 8. But for 26, sum is 8, not divisible by 8. This rule is incorrect for 8.

(B) Its last digit is divisible by 8. This is incorrect. For example, the last digit of 18 is 8, which is divisible by 8, but 18 is not divisible by 8. The last digit of 24 is 4, which is not divisible by 8, but 24 is divisible by 8. This rule is incorrect for 8.

(C) The number formed by its last two digits is divisible by 8. This is the divisibility rule for 4. It is not the rule for 8. For example, the last two digits of 112 are 12, which is divisible by 4, and 112 is divisible by 8 ($112 \div 8 = 14$). However, the last two digits of 212 are 12, which is divisible by 4, but 212 is not divisible by 8 ($212 \div 8 = 26$ with a remainder of 4). This rule is incorrect for 8.

(D) The number formed by its last three digits is divisible by 8. This is the correct divisibility rule for 8. For example, consider the number 3120. The number formed by the last three digits is 120. $120 \div 8 = 15$. Since 120 is divisible by 8, 3120 is divisible by 8 ($3120 \div 8 = 390$). Consider the number 123456. The number formed by the last three digits is 456. $456 \div 8 = 57$. Since 456 is divisible by 8, 123456 is divisible by 8.

This statement is correct.

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The correct divisibility rule for 8 is that the number formed by its last three digits must be divisible by 8.

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The correct option is (D) The number formed by its last three digits is divisible by 8.

The Highest Common Factor (HCF) of three numbers is the largest positive integer that divides each of the three numbers exactly.

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We can find the HCF by using the prime factorization method.

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Find the prime factorization of each number:

Prime factorization of 20:

$20 = 2 \times 10 = 2 \times 2 \times 5 = 2^2 \times 5^1$

Prime factorization of 28:

$28 = 2 \times 14 = 2 \times 2 \times 7 = 2^2 \times 7^1$

Prime factorization of 36:

$36 = 2 \times 18 = 2 \times 2 \times 9 = 2^2 \times 3 \times 3 = 2^2 \times 3^2$

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To find the HCF, we identify the common prime factors and take the lowest power of each common factor.

The prime factors present are 2, 3, 5, and 7.

The only prime factor common to all three numbers (20, 28, and 36) is 2.

The powers of the common prime factor 2 are $2^2$ (in 20), $2^2$ (in 28), and $2^2$ (in 36).

The lowest power of the common factor 2 is $2^2$.

The prime factors 3, 5, and 7 are not common to all three numbers.

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HCF$(20, 28, 36) = 2^2 = 4$.

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The HCF of 20, 28, and 36 is 4.

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The correct option is (B) 4.

The Least Common Multiple (LCM) of three numbers is the smallest positive integer that is a multiple of all three numbers.

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We can find the LCM by using the prime factorization method.

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Find the prime factorization of each number:

Prime factorization of 20:

$20 = 2 \times 10 = 2 \times 2 \times 5 = 2^2 \times 5^1$

Prime factorization of 28:

$28 = 2 \times 14 = 2 \times 2 \times 7 = 2^2 \times 7^1$

Prime factorization of 36:

$36 = 2 \times 18 = 2 \times 2 \times 9 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$

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To find the LCM, we take the highest power of all prime factors that appear in the factorization of any of the numbers and multiply them together.

The prime factors present are 2, 3, 5, and 7.

Highest power of 2 is $2^2$ (from 20, 28, and 36).

Highest power of 3 is $3^2$ (from 36).

Highest power of 5 is $5^1$ (from 20).

Highest power of 7 is $7^1$ (from 28).

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LCM$(20, 28, 36) = 2^2 \times 3^2 \times 5^1 \times 7^1$

LCM $= 4 \times 9 \times 5 \times 7$

LCM $= 36 \times 35$

Performing the multiplication:

$\begin{array}{cc}& & 3 & 6 \\ \times & & 3 & 5 \\ \hline & 1 & 8 & 0 \\ 1 & 0 & 8 & \times \\ \hline 1 & 2 & 6 & 0 \\ \hline \end{array}$

LCM = 1260.

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The LCM of 20, 28, and 36 is 1260.

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The correct option is (A) 1260.

We are asked to find the value of the sum of two prime numbers, 2 and 7.

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The prime numbers given are 2 and 7.

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Let's calculate their sum:

$2 + 7 = 9$

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The result of the addition is 9.

Now, let's look at the options:

(A) 9: This is the value of the sum.

(B) 10: This is not the value of the sum.

(C) An even number: 9 is an odd number, not an even number.

(D) A composite number: A composite number is a natural number greater than 1 that has more than two distinct positive divisors. The factors of 9 are 1, 3, and 9. Since 9 has factors other than 1 and itself, it is a composite number. While 9 is a composite number, the blank asks for the value of the sum, not a property of the sum.

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The blank should be filled with the result of the addition, which is 9.

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The correct option is (A) 9.

A factor of a number is a number that divides the given number exactly, leaving no remainder.

A natural number greater than 1 that has exactly two distinct positive factors (1 and itself) is called a prime number.

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We need to find which of the given numbers has exactly two factors by listing the positive factors for each option.

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(A) 1: The positive factors of 1 are only 1. It has exactly one factor.

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(B) 2: The positive factors of 2 are 1 and 2. These are the only numbers that divide 2 exactly. It has exactly two factors.

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(C) 7: The positive factors of 7 are 1 and 7. These are the only numbers that divide 7 exactly. It has exactly two factors.

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(D) 10: The positive factors of 10 are 1, 2, 5, and 10. These are the numbers that divide 10 exactly. It has four factors.

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Based on the number of factors:

1 has 1 factor.

2 has 2 factors.

7 has 2 factors.

10 has 4 factors.

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The numbers that have exactly two factors are 2 and 7.

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The correct options are (B) 2 and (C) 7.

Given:

Time taken by Amit for one round = 12 minutes

Time taken by Binod for one round = 15 minutes

Time taken by Charan for one round = 20 minutes

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To Find:

The time after which they will meet again at the starting point for the first time.

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Solution:

To find the time when they will all meet again at the starting point for the first time, we need to find the Least Common Multiple (LCM) of their individual times taken to complete one round.

We need to find the LCM of 12, 15, and 20.

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Let's find the prime factorization of each number:

Prime factorization of 12:

$12 = 2 \times 2 \times 3 = 2^2 \times 3^1$

Prime factorization of 15:

$15 = 3 \times 5 = 3^1 \times 5^1$

Prime factorization of 20:

$20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

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

The unique prime factors are 2, 3, and 5.

Highest power of 2: $2^2$ (from 12 and 20)

Highest power of 3: $3^1$ (from 12 and 15)

Highest power of 5: $5^1$ (from 15 and 20)

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

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The LCM of 12, 15, and 20 is 60.

This means that after 60 minutes, Amit will have completed $60 \div 12 = 5$ rounds, Binod will have completed $60 \div 15 = 4$ rounds, and Charan will have completed $60 \div 20 = 3$ rounds. Since each has completed a whole number of rounds, they will all be back at the starting point simultaneously.

This is the first time they will meet again at the starting point after they started.

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The correct option is (B) 60 minutes.

We are given that a number is divisible by both 2 and 5.

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A number is divisible by a composite number if it is divisible by its coprime factors.

The numbers 2 and 5 are prime numbers, and they are coprime because their greatest common divisor is 1 ($\text{HCF}(2, 5) = 1$).

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If a number is divisible by two coprime numbers, then it is also divisible by their product.

The product of 2 and 5 is $2 \times 5 = 10$.

Therefore, if a number is divisible by both 2 and 5, it must be divisible by 10.

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Let's verify with an example. Consider the number 20. 20 is divisible by 2 ($20 \div 2 = 10$) and 20 is divisible by 5 ($20 \div 5 = 4$). 20 is also divisible by 10 ($20 \div 10 = 2$).

Consider the number 30. 30 is divisible by 2 ($30 \div 2 = 15$) and 30 is divisible by 5 ($30 \div 5 = 6$). 30 is also divisible by 10 ($30 \div 10 = 3$).

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Based on the divisibility rule and examples, a number divisible by both 2 and 5 is divisible by 10.

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Let's look at the options:

(A) 7: Not necessarily. 10 is divisible by 2 and 5, but not by 7.

(B) 10: Yes, as determined by the divisibility rule.

(C) 15: Not necessarily. 10 is divisible by 2 and 5, but not by 15.

(D) 20: Not necessarily. 10 is divisible by 2 and 5, but not by 20.

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The correct option is (B) 10.

Prime factorization is the process of expressing a composite number as a product of its prime factors.

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We need to find the prime numbers that multiply together to equal 100.

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Let's find the prime factors of 100:

Start by dividing 100 by the smallest prime number, which is 2:

$100 \div 2 = 50$

Divide 50 by 2 again:

$50 \div 2 = 25$

Now 25 is not divisible by 2. The next prime number is 3. The sum of digits of 25 is $2+5=7$, which is not divisible by 3, so 25 is not divisible by 3.

The next prime number is 5. 25 is divisible by 5:

$25 \div 5 = 5$

5 is a prime number. Divide 5 by 5:

$5 \div 5 = 1$

We stop when we reach 1.

The prime factors of 100 are 2, 2, 5, and 5.

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Writing 100 as a product of its prime factors:

$100 = 2 \times 2 \times 5 \times 5$

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Using exponents to represent the repeated factors:

$100 = 2^2 \times 5^2$

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Now let's examine the given options:

(A) $2^2 \times 5^2$: This is $(2 \times 2) \times (5 \times 5) = 4 \times 25 = 100$. The factors 2 and 5 are prime. This is the prime factorization.

(B) $2 \times 50$: 50 is not a prime number ($50 = 2 \times 5 \times 5$). This is not the prime factorization.

(C) $4 \times 25$: Neither 4 ($4=2^2$) nor 25 ($25=5^2$) is a prime number. This is not the prime factorization.

(D) $10 \times 10$: 10 is not a prime number ($10=2 \times 5$). This is not the prime factorization.

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The prime factorization of 100 is $2^2 \times 5^2$.

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The correct option is (A) $2^2 \times 5^2$.

Co-prime numbers (or relatively prime numbers) are two integers whose only common positive divisor is 1. In other words, their Highest Common Factor (HCF) is 1.

We need to find the pair of numbers among the given options whose HCF is not 1.

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Let's find the HCF for each pair:

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(A) (8, 9):

Factors of 8: 1, 2, 4, 8

Factors of 9: 1, 3, 9

Common factor is 1. HCF(8, 9) = 1. This is a pair of co-prime numbers.

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(B) (15, 16):

Factors of 15: 1, 3, 5, 15

Factors of 16: 1, 2, 4, 8, 16

Common factor is 1. HCF(15, 16) = 1. This is a pair of co-prime numbers.

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(C) (12, 18):

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

Factors of 18: 1, 2, 3, 6, 9, 18

Common factors are 1, 2, 3, and 6. The highest common factor is 6. HCF(12, 18) = 6. Since the HCF is 6 (not 1), this pair is NOT co-prime.

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(D) (21, 25):

Factors of 21: 1, 3, 7, 21

Factors of 25: 1, 5, 25

Common factor is 1. HCF(21, 25) = 1. This is a pair of co-prime numbers.

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The pair that does not have 1 as its only common positive divisor is (12, 18).

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The correct option is (C) (12, 18).

Given:

Product of two numbers = 120

HCF of the two numbers = 4

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To Find:

LCM of the two numbers.

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Solution:

We use the fundamental relationship between the HCF and LCM of two positive integers:

Product of the two numbers = HCF $\times$ LCM

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Let the two numbers be $a$ and $b$. We are given $a \times b = 120$ and $\text{HCF}(a, b) = 4$. We need to find $\text{LCM}(a, b)$.

Substitute the given values into the formula:

$120 = 4 \times \text{LCM}(a, b)$

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Now, we solve for the LCM:

$\text{LCM}(a, b) = \frac{120}{4}$

$\text{LCM}(a, b) = 30$

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The LCM of the two numbers is 30.

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The correct option is (C) 30.

The smallest 3-digit number is 100.

A number is divisible by 6 if it is divisible by both 2 and 3.

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We need to find the smallest number greater than or equal to 100 that is divisible by both 2 and 3.

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Let's check the numbers starting from 100:

100: Last digit is 0, so divisible by 2. Sum of digits = $1+0+0=1$. 1 is not divisible by 3. So, 100 is not divisible by 6.

101: Last digit is 1, so not divisible by 2. Not divisible by 6.

102: Last digit is 2, so divisible by 2. Sum of digits = $1+0+2=3$. 3 is divisible by 3. So, 102 is divisible by both 2 and 3.

Since 102 is divisible by both 2 and 3, it is divisible by 6.

$102 \div 6 = 17$.

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Since 102 is the first 3-digit number we encountered that is divisible by 6, it is the smallest 3-digit number divisible by 6.

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Let's check the other options to confirm:

(C) 108: Last digit is 8 (divisible by 2). Sum of digits $1+0+8=9$ (divisible by 3). 108 is divisible by 6. But 108 is larger than 102.

(D) 112: Last digit is 2 (divisible by 2). Sum of digits $1+1+2=4$ (not divisible by 3). 112 is not divisible by 6.

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The smallest 3-digit number divisible by 6 is 102.

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The correct option is (B) 102.

A factor of a number is a number that divides into it exactly, leaving no remainder.

Example:

The factors of 12 are the numbers that divide 12 without a remainder. These are 1, 2, 3, 4, 6, and 12.

We can check this by division:

$12 \div 1 = 12$

$12 \div 2 = 6$

$12 \div 3 = 4$

$12 \div 4 = 3$

$12 \div 6 = 2$

$12 \div 12 = 1$

Since these divisions result in whole numbers with no remainder, 1, 2, 3, 4, 6, and 12 are all factors of 12.

A multiple of a number is the result of multiplying that number by an integer (typically a positive integer when referring to listing multiples, but the definition includes zero and negative integers).

In simpler terms, the multiples of a number are the numbers that appear in its multiplication table.

Example:

Let's find the multiples of 7.

We multiply 7 by integers:

$7 \times 1 = 7$

$7 \times 2 = 14$

$7 \times 3 = 21$

$7 \times 4 = 28$

...and so on.

So, the multiples of 7 are 7, 14, 21, 28, 35, 42, ...

Note that 0 ($7 \times 0 = 0$) is also a multiple of any number, and negative numbers like -7, -14, etc. are also multiples.

To find all the factors of 36, we look for pairs of numbers that multiply together to give 36.

We can start by dividing 36 by small integers 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)

$36 \div 5$ does not divide exactly.

$36 \div 6 = 6$ (So, 6 is a factor. We stop here because the quotient, 6, is equal to the divisor, 6).

Listing all the unique divisors we found gives us the factors.

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

To find the first five multiples of 11, we multiply 11 by the first five positive integers (1, 2, 3, 4, and 5).

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$11 \times 1 = 11$

$11 \times 2 = 22$

$11 \times 3 = 33$

$11 \times 4 = 44$

$11 \times 5 = 55$

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The first five multiples of 11 are 11, 22, 33, 44, and 55.

To determine if 8 is a factor of 64, we need to see if 64 is divisible by 8 without leaving a remainder.

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We perform the division:

$64 \div 8$

We know from multiplication tables that $8 \times 8 = 64$.

Therefore, $64 \div 8 = 8$.

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Since the division of 64 by 8 results in a whole number (8) with no remainder, 8 is a factor of 64.

Yes, 8 is a factor of 64 because 64 is perfectly divisible by 8.

To determine if 75 is a multiple of 8, we need to see if 75 can be obtained by multiplying 8 by an integer. This is equivalent to checking if 75 is divisible by 8 without a remainder.

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Let's perform the division $75 \div 8$:

$75 = 8 \times 9 + 3$

When we divide 75 by 8, we get a quotient of 9 and a remainder of 3.

Since there is a remainder (3 is not 0), 75 is not perfectly divisible by 8.

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Therefore, 75 is not a multiple of 8.

No, 75 is not a multiple of 8 because 75 cannot be expressed as $8 \times \text{an integer}$.

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

In other words, a prime number has exactly two distinct positive factors: 1 and the number itself.

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Examples:

1. The number 2 is a prime number. Its only positive factors are 1 and 2.

2. The number 3 is a prime number. Its only positive factors are 1 and 3.

3. The number 5 is a prime number. Its only positive factors are 1 and 5.

4. The number 7 is a prime number. Its only positive factors are 1 and 7.

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Note: The number 1 is not considered a prime number as it only has one positive factor (itself).

A composite number is a natural number greater than 1 that is not prime.

In other words, a composite number is a natural number greater than 1 that has at least one positive divisor other than 1 and itself.

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Examples:

1. The number 4 is a composite number. Its positive factors are 1, 2, and 4 (it has a factor other than 1 and itself, which is 2).

2. The number 6 is a composite number. Its positive factors are 1, 2, 3, and 6 (it has factors other than 1 and itself, such as 2 and 3).

3. The number 8 is a composite number. Its positive factors are 1, 2, 4, and 8.

4. The number 9 is a composite number. Its positive factors are 1, 3, and 9.

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Note: The number 1 is neither prime nor composite.

The number 1 is neither a prime number nor a composite number.

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Here's the explanation:

According to the definition, a prime number is a natural number greater than 1 that has exactly two distinct positive factors (1 and itself).

The number 1 has only one positive factor, which is 1. It does not have two distinct factors, and it is not greater than 1.

A composite number is a natural number greater than 1 that has more than two positive factors.

The number 1 is not greater than 1 and has only one factor.

Since 1 does not fit the definition of either a prime number or a composite number, it is classified as neither.

To list the prime numbers between 1 and 20, we consider each number in this range (excluding 1) and check if it has exactly two positive factors (1 and itself).

The numbers between 1 and 20 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.

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Let's check each number:

2: Factors are 1, 2. (Prime)

3: Factors are 1, 3. (Prime)

4: Factors are 1, 2, 4. (Composite)

5: Factors are 1, 5. (Prime)

6: Factors are 1, 2, 3, 6. (Composite)

7: Factors are 1, 7. (Prime)

8: Factors are 1, 2, 4, 8. (Composite)

9: Factors are 1, 3, 9. (Composite)

10: Factors are 1, 2, 5, 10. (Composite)

11: Factors are 1, 11. (Prime)

12: Factors are 1, 2, 3, 4, 6, 12. (Composite)

13: Factors are 1, 13. (Prime)

14: Factors are 1, 2, 7, 14. (Composite)

15: Factors are 1, 3, 5, 15. (Composite)

16: Factors are 1, 2, 4, 8, 16. (Composite)

17: Factors are 1, 17. (Prime)

18: Factors are 1, 2, 3, 6, 9, 18. (Composite)

19: Factors are 1, 19. (Prime)

20: Factors are 1, 2, 4, 5, 10, 20. (Composite)

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The prime numbers between 1 and 20 are 2, 3, 5, 7, 11, 13, 17, and 19.

Twin prime numbers are pairs of prime numbers that differ by 2.

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Example:

The pair of numbers (3, 5) are twin primes because both 3 and 5 are prime numbers, and their difference is $5 - 3 = 2$.

Other examples of twin prime pairs include:

(5, 7)

(11, 13)

(17, 19)

(29, 31)

Two numbers are considered co-prime (or relatively prime) if their greatest common divisor (GCD) is 1. This means the only positive integer that divides both numbers is 1.

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To check if 15 and 16 are co-prime, we need to find their factors.

Factors of 15: 1, 3, 5, 15

Factors of 16: 1, 2, 4, 8, 16

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Now, let's identify the common factors:

The factors that appear in both lists are only 1.

The greatest common divisor (GCD) of 15 and 16 is 1.

---

Since the GCD of 15 and 16 is 1, the numbers 15 and 16 are indeed co-prime numbers.

Yes, 15 and 16 are co-prime because their only common positive factor is 1.

The divisibility rule for 2 states that a number is divisible by 2 if its last digit is an even number (0, 2, 4, 6, or 8).

---

To check if 5678 is divisible by 2, we look at its last digit.

The last digit of 5678 is 8.

Since 8 is an even number, according to the divisibility rule, 5678 is divisible by 2.

---

Yes, 5678 is divisible by 2 because its last digit is 8, which is an even number.

The divisibility rule for 5 states that a number is divisible by 5 if its last digit is either 0 or 5.

---

To check if 9035 is divisible by 5, we look at its last digit.

The last digit of 9035 is 5.

Since the last digit is 5, according to the divisibility rule, 9035 is divisible by 5.

---

Yes, 9035 is divisible by 5 because its last digit is 5.

The divisibility rule for 10 states that a number is divisible by 10 if its last digit is 0.

---

To check if 780 is divisible by 10, we look at its last digit.

The last digit of 780 is 0.

Since the last digit is 0, according to the divisibility rule, 780 is divisible by 10.

---

Yes, 780 is divisible by 10 because its last digit is 0.

The divisibility rule for 3 states that a number is divisible by 3 if the sum of its digits is divisible by 3.

---

To check if 4321 is divisible by 3, we find the sum of its digits:

Sum of digits $= 4 + 3 + 2 + 1 = 10$

---

Now, we check if the sum of the digits (10) is divisible by 3.

$10 \div 3$ leaves a remainder of 1. 10 is not divisible by 3.

Since the sum of the digits (10) is not divisible by 3, according to the divisibility rule, 4321 is not divisible by 3.

---

No, 4321 is not divisible by 3 because the sum of its digits ($4+3+2+1=10$) is not divisible by 3.

The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is divisible by 9.

---

To check if 6543 is divisible by 9, we find the sum of its digits:

Sum of digits $= 6 + 5 + 4 + 3 = 18$

---

Now, we check if the sum of the digits (18) is divisible by 9.

$18 \div 9 = 2$. 18 is divisible by 9.

Since the sum of the digits (18) is divisible by 9, according to the divisibility rule, 6543 is divisible by 9.

---

Yes, 6543 is divisible by 9 because the sum of its digits ($6+5+4+3=18$) is divisible by 9.

Prime factorisation is the process of finding the prime numbers that multiply together to make the original number.

Every composite number can be expressed as a unique product of prime numbers.

---

To find the prime factorisation of 54, we can use the division method:

We divide 54 by the smallest prime number that divides it exactly, and continue the process until we reach 1.

$\begin{array}{c|cc} 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

---

From the division, we see that $54 = 2 \times 3 \times 3 \times 3$.

The prime factors of 54 are 2 and 3.

The prime factorisation of 54 is $2 \times 3^3$.

The Highest Common Factor (HCF) of two or more numbers is the largest number that is a factor of all of them.

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To find the HCF of 15 and 25 by listing factors:

First, list all the factors of 15:

Factors of 15: 1, 3, 5, 15

---

Next, list all the factors of 25:

Factors of 25: 1, 5, 25

---

Now, identify the common factors (factors that are in both lists):

Common factors of 15 and 25 are 1 and 5.

---

The highest (largest) of these common factors is 5.

Therefore, the HCF of 15 and 25 is 5.

The Least Common Multiple (LCM) of two or more numbers is the smallest positive number that is a multiple of all of them.

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To find the LCM of 8 and 12 by listing multiples:

First, list the multiples of 8:

Multiples of 8: 8, 16, 24, 32, 40, 48, 56, ...

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Next, list the multiples of 12:

Multiples of 12: 12, 24, 36, 48, 60, ...

---

Now, identify the common multiples (multiples that appear in both lists):

Common multiples of 8 and 12 are 24, 48, ...

---

The least (smallest) of these common multiples is 24.

Therefore, the LCM of 8 and 12 is 24.

Yes, if a number is divisible by 6, it is necessarily divisible by both 2 and 3.

---

Here's the explanation:

Divisibility by a composite number depends on its prime factors.

The number 6 is a composite number, and its prime factorisation is $2 \times 3$.

If a number is divisible by 6, it means that 6 is a factor of that number. For 6 to be a factor, the number must contain both 2 and 3 as prime factors in its own prime factorisation.

For example, consider the number 18.

18 is divisible by 6 ($18 \div 6 = 3$).

The prime factorisation of 18 is $2 \times 3 \times 3$. Notice that the prime factors 2 and 3 are present.

Since the prime factors of 6 (which are 2 and 3) are present in the prime factorisation of any number divisible by 6, that number must also be divisible by 2 and by 3 individually.

No, the Highest Common Factor (HCF) of two distinct numbers cannot be greater than the smaller number.

---

Here's the reason:

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

By definition, any factor of a number must be less than or equal to that number.

Since the HCF is a factor of both the given numbers, it must be less than or equal to the smaller of the two numbers.

If the HCF were greater than the smaller number, it would mean that the HCF is a factor of the smaller number and is also greater than the smaller number, which is impossible for positive integers.

For example, the smaller number is 10. Any factor of 10 must be $\leq 10$ (the factors are 1, 2, 5, 10). The HCF of 10 and any other number (say 15) must be a factor of 10, so it cannot be greater than 10. (HCF of 10 and 15 is 5, which is $\leq 10$).

No, the Least Common Multiple (LCM) of two distinct numbers cannot be smaller than the larger number.

---

Here's the reason:

The LCM of two numbers is the smallest positive number that is a multiple of both of them.

By definition, any multiple of a number must be greater than or equal to that number (for positive multiples).

Since the LCM is a multiple of both the given numbers, it must be greater than or equal to the larger of the two numbers.

If the LCM were smaller than the larger number, it would mean that the LCM is a multiple of the larger number and is also smaller than the larger number, which is impossible for positive integers.

For example, let the larger number be 12. Any multiple of 12 must be $\geq 12$ (the positive multiples are 12, 24, 36, ...). The LCM of 12 and any other distinct number (say 8) must be a multiple of 12, so it cannot be smaller than 12. (LCM of 8 and 12 is 24, which is $\geq 12$).

If the Highest Common Factor (HCF) of two numbers is 1, it means that the only positive integer that divides both numbers exactly is 1.

Such numbers are called co-prime or relatively prime numbers.

---

Are they necessarily prime?

No, the numbers are not necessarily prime just because their HCF is 1.

---

Here's why:

Co-prime numbers only need to share 1 as a common factor. They themselves can be composite numbers.

Examples:

1. Consider the numbers 4 and 9.

Factors of 4: 1, 2, 4

Factors of 9: 1, 3, 9

The common factor is only 1. So, HCF(4, 9) = 1.

Both 4 and 9 are composite numbers.

2. Consider the numbers 8 and 15.

Factors of 8: 1, 2, 4, 8

Factors of 15: 1, 3, 5, 15

The common factor is only 1. So, HCF(8, 15) = 1.

Both 8 and 15 are composite numbers.

Therefore, while prime numbers are always co-prime to any other number except their own multiples, composite numbers can also be co-prime to other composite numbers or prime numbers.

Let the two numbers be $a$ and $b$.

We are given:

HCF of the two numbers = 4

LCM of the two numbers = 48

One number, let's say $a$, is 16.

We need to find the other number, $b$.

---

There is a fundamental relationship between the HCF, LCM, and the product of two numbers:

Product of the two numbers = HCF $\times$ LCM

Mathematically, this is expressed as:

$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$

---

Now, we substitute the given values into this relationship:

$16 \times b = 4 \times 48$

---

Calculate the product on the right side:

$4 \times 48 = 192$

So, the equation becomes:

$16 \times b = 192$

---

To find $b$, divide both sides of the equation by 16:

$b = \frac{192}{16}$

---

Performing the division:

$192 \div 16 = 12$

So, $b = 12$.

---

The other number is 12.

The key difference between factors and multiples lies in their relationship to the given number through division and multiplication.

---

Factors of a number:

A factor of a number is a number that divides into it exactly, leaving no remainder. Factors are typically less than or equal to the number itself (excluding the number itself, other factors are less than it).

To find the factors of 18, we look for numbers that divide 18 evenly:

$18 \div 1 = 18$

$18 \div 2 = 9$

$18 \div 3 = 6$

$18 \div 6 = 3$

$18 \div 9 = 2$

$18 \div 18 = 1$

The positive factors of 18 are the divisors: 1, 2, 3, 6, 9, and 18.

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Multiples of a number:

A multiple of a number is the result of multiplying that number by an integer. Multiples are typically greater than or equal to the number itself (for positive integer multipliers).

To find the first 7 positive multiples of 18, we multiply 18 by the integers 1, 2, 3, 4, 5, 6, and 7:

$18 \times 1 = 18$

$18 \times 2 = 36$

$18 \times 3 = 54$

$18 \times 4 = 72$

$18 \times 5 = 90$

$18 \times 6 = 108$

$18 \times 7 = 126$

The first 7 multiples of 18 are: 18, 36, 54, 72, 90, 108, and 126.

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In summary, factors divide the number, while multiples are products of the number.

Prime numbers are natural numbers greater than 1 that have exactly two distinct positive factors: 1 and the number itself.

Composite numbers are natural numbers greater than 1 that have more than two positive factors.

The number 1 is neither prime nor composite.

---

Method to identify if a number is prime or composite:

To determine if a number $n > 1$ is prime or composite, we can test its divisibility by prime numbers starting from 2, 3, 5, 7, and so on, up to the square root of $n$ ($\sqrt{n}$).

If the number $n$ is not divisible by any prime number less than or equal to its square root, then $n$ is a prime number.

If the number $n$ is divisible by any prime number less than or equal to its square root, then $n$ is a composite number.

This method works because if $n$ has a composite factor greater than $\sqrt{n}$, then it must also have a prime factor smaller than $\sqrt{n}$. If $n$ has two prime factors greater than $\sqrt{n}$, their product would be greater than $(\sqrt{n})^2 = n$, which is impossible if they are factors of $n$. Thus, any composite number $n$ must have at least one prime factor less than or equal to $\sqrt{n}$.

---

Example 1: Is 97 prime or composite?

First, find the approximate square root of 97:

We know that $9^2 = 81$ and $10^2 = 100$. So, $\sqrt{97}$ is between 9 and 10. We only need to test prime divisors up to 9.

The prime numbers less than or equal to 9 are 2, 3, 5, and 7.

Now, we test the divisibility of 97 by these primes:

Divisibility by 2: The last digit of 97 is 7 (odd). So, 97 is not divisible by 2.

Divisibility by 3: Sum of digits = $9 + 7 = 16$. 16 is not divisible by 3. So, 97 is not divisible by 3.

Divisibility by 5: The last digit of 97 is 7 (not 0 or 5). So, 97 is not divisible by 5.

Divisibility by 7: $97 \div 7$. $7 \times 10 = 70$, $97 - 70 = 27$. $7 \times 3 = 21$, $7 \times 4 = 28$. 97 is not exactly divisible by 7 (remainder 6). So, 97 is not divisible by 7.

Since 97 is not divisible by any prime number less than or equal to its square root, 97 is a prime number.

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Example 2: Is 111 prime or composite?

First, find the approximate square root of 111:

We know that $10^2 = 100$ and $11^2 = 121$. So, $\sqrt{111}$ is between 10 and 11. We only need to test prime divisors up to 10.

The prime numbers less than or equal to 10 are 2, 3, 5, and 7.

Now, we test the divisibility of 111 by these primes:

Divisibility by 2: The last digit of 111 is 1 (odd). So, 111 is not divisible by 2.

Divisibility by 3: Sum of digits = $1 + 1 + 1 = 3$. 3 is divisible by 3. So, 111 is divisible by 3.

Since 111 is divisible by 3 (which is a prime number less than its square root), 111 is a composite number.

Note: $111 = 3 \times 37$. Both 3 and 37 are prime factors.

Here are the divisibility rules for 4, 6, and 8 and their application to the number 28764.

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Divisibility Rule for 4:

A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

Explanation: The number 100 is divisible by 4 ($100 \div 4 = 25$). Any number can be written as $100 \times (\text{some integer}) + (\text{number formed by last two digits})$. Since $100 \times (\text{some integer})$ is always divisible by 4, the divisibility of the entire number by 4 depends solely on the divisibility of the number formed by its last two digits by 4.

Check for 28764:

The number formed by the last two digits of 28764 is 64.

Is 64 divisible by 4? Yes, $64 \div 4 = 16$.

Therefore, 28764 is divisible by 4.

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Divisibility Rule for 6:

A number is divisible by 6 if it is divisible by both 2 and 3.

Explanation: Since 6 is the product of two prime numbers, 2 and 3, a number must have both 2 and 3 as factors to be divisible by 6. We check for divisibility by 2 and 3 using their respective rules.

Check for 28764:

Divisibility by 2: The last digit of 28764 is 4 (even). So, 28764 is divisible by 2.

Divisibility by 3: Sum of digits = $2 + 8 + 7 + 6 + 4 = 27$. Is 27 divisible by 3? Yes, $27 \div 3 = 9$. So, 28764 is divisible by 3.

Since 28764 is divisible by both 2 and 3, it is divisible by 6.

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Divisibility Rule for 8:

A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

Explanation: The number 1000 is divisible by 8 ($1000 \div 8 = 125$). Any number greater than or equal to 1000 can be written as $1000 \times (\text{some integer}) + (\text{number formed by last three digits})$. Since $1000 \times (\text{some integer})$ is always divisible by 8, the divisibility of the entire number by 8 depends solely on the divisibility of the number formed by its last three digits by 8.

Check for 28764:

The number formed by the last three digits of 28764 is 764.

Is 764 divisible by 8?

$764 \div 8 = 95$ with a remainder of 4 ($8 \times 95 = 760$, $764 - 760 = 4$).

Since 764 is not divisible by 8, 28764 is not divisible by 8.

Here are the divisibility rules for 9 and 11 and their application to the number 50719.

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Divisibility Rule for 9:

A number is divisible by 9 if the sum of its digits is divisible by 9.

Explanation: This rule is similar to the divisibility rule for 3. It is based on the fact that any power of 10 leaves a remainder of 1 when divided by 9 (e.g., $10 \div 9 = 1$ remainder 1, $100 \div 9 = 11$ remainder 1, etc.). When you write a number in expanded form (e.g., $50719 = 5 \times 10000 + 0 \times 1000 + 7 \times 100 + 1 \times 10 + 9 \times 1$), the remainders when divided by 9 are equivalent to the sum of the digits.

Check for 50719:

Sum of digits $= 5 + 0 + 7 + 1 + 9 = 22$

Is 22 divisible by 9? No, $22 \div 9 = 2$ with a remainder of 4.

Therefore, 50719 is not divisible by 9.

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Divisibility Rule for 11:

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

Explanation: This rule relies on the pattern of remainders when powers of 10 are divided by 11 ($10^0 = 1$, $10^1 = 10 \equiv -1 \pmod{11}$, $10^2 = 100 \equiv 1 \pmod{11}$, $10^3 = 1000 \equiv -1 \pmod{11}$, and so on). The remainders alternate between 1 and -1. When you write a number in expanded form and consider its remainder modulo 11, the digits at odd places (from the right) are multiplied by powers of 10 with an even exponent (leaving remainder 1), and digits at even places (from the right) are multiplied by powers of 10 with an odd exponent (leaving remainder -1). This leads to the alternating sum/difference pattern.

Check for 50719:

Let's write the number and label the digit positions from the right:

$5\quad 0\quad 7\quad 1\quad 9$

Pos: 5 4 3 2 1 (from right)

Odd places (1st, 3rd, 5th): Digits are 9, 7, 5

Sum of digits at odd places $= 9 + 7 + 5 = 21$

Even places (2nd, 4th): Digits are 1, 0

Sum of digits at even places $= 1 + 0 = 1$

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

Difference $= 21 - 1 = 20$

Is 20 equal to 0 or a multiple of 11? No, 20 is not 0, and $20 \div 11$ leaves a remainder (e.g., $11 \times 1 = 11$, $11 \times 2 = 22$).

Therefore, 50719 is not divisible by 11.

Prime factorisation is the process of breaking down a composite number into its prime factors, which, when multiplied together, give the original number. Every composite number has a unique prime factorization.

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Method 1: Factor Tree Method

In the factor tree method, we start with the given number at the top and draw branches below it, representing a pair of factors whose product is the number. We continue factoring composite numbers in this way until all the numbers at the ends of the branches are prime. The prime numbers at the ends of the branches are the prime factors.

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Prime Factorisation of 210 using Factor Tree:

We can start with any pair of factors, for example, 10 and 21.

210

/ \ \ \\

10 21

/ \ / \ \\

2 5 3 7

Since 2, 5, 3, and 7 are all prime numbers, we stop here.

The prime factors are 2, 3, 5, and 7.

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Prime Factorisation of 432 using Factor Tree:

We can start with any pair of factors, for example, 2 and 216.

432

/ \ \ \\

2 216

\ / \ \\

2 108

\ / \ \ \\

2 54

\ / \ \ \\

2 27

\ / \ \ \\

3 9

\ / \ \\

3 3

The prime factors are 2 (four times) and 3 (three times).

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Method 2: Division Method

In the division method, we divide the number by the smallest prime number that divides it exactly. We then divide the quotient by the smallest prime number that divides it exactly, and continue this process until the quotient is 1. The prime divisors used are the prime factors of the original number.

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Prime Factorisation of 210 using Division Method:

$\begin{array}{c|cc} 2 & 210 \\ \hline 3 & 105 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

The prime factors are 2, 3, 5, and 7.

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Prime Factorisation of 432 using Division Method:

$\begin{array}{c|cc} 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factors are 2 (four times) and 3 (three times).

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Summary:

The prime factorisation of 210 is $2 \times 3 \times 5 \times 7$.

The prime factorisation of 432 is $2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^4 \times 3^3$.

The Highest Common Factor (HCF) of two or more numbers is the largest number that divides into all of them exactly.

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Method to find HCF using Prime Factorisation:

1. Find the prime factorisation of each number.

2. Identify the prime factors that are common to all the numbers.

3. For each common prime factor, take the lowest power that appears in the factorisations.

4. Multiply these lowest powers of the common prime factors together. The result is the HCF.

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Find the HCF of 120 and 144 using Prime Factorisation:

Step 1: Find the prime factorisation of 120 and 144.

Prime factorisation of 120:

$\begin{array}{c|cc} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3^1 \times 5^1$

Prime factorisation of 144:

$\begin{array}{c|cc} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$

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Step 2: Identify common prime factors.

The prime factors common to both 120 and 144 are 2 and 3.

---

Step 3: Take the lowest power of each common prime factor.

For the prime factor 2: The powers are $2^3$ (in 120) and $2^4$ (in 144). The lowest power is $2^3$.

For the prime factor 3: The powers are $3^1$ (in 120) and $3^2$ (in 144). The lowest power is $3^1$.

The prime factor 5 is not common to both, so it is not included in the HCF.

---

Step 4: Multiply the lowest powers.

HCF(120, 144) $= 2^3 \times 3^1 = (2 \times 2 \times 2) \times 3 = 8 \times 3 = 24$

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The HCF of 120 and 144 is 24.

The Least Common Multiple (LCM) of two or more numbers is the smallest positive number that is a multiple of all of them.

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Method to find LCM using Common Division Method:

1. Write the given numbers in a row, separated by commas.

2. Divide the numbers by the smallest prime number that divides at least one of the numbers exactly. Write the quotients below the respective numbers. If a number is not divisible by the prime, write the number as it is in the next row.

3. Repeat the process with the new quotients until all the numbers in the last row are 1.

4. The LCM is the product of all the prime divisors used in the process.

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Find the LCM of 90 and 108 using Common Division Method:

We arrange 90 and 108 and divide by common prime factors, then by remaining factors:

$\begin{array}{c|cc} 2 & 90 \;, & 108 \\ \hline 3 & 45 \; , & 54 \\ \hline 3 & 15 \; , & 18 \\ \hline 2 & 5 \; , & 6 \\ \hline 5 & 5 \; , & 3 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$

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The prime divisors used are 2, 3, 3, 2, 5, and 3.

To find the LCM, we multiply all these divisors:

LCM(90, 108) $= 2 \times 3 \times 3 \times 2 \times 5 \times 3$

LCM(90, 108) $= (2 \times 2) \times (3 \times 3 \times 3) \times 5$

LCM(90, 108) $= 4 \times 27 \times 5$

LCM(90, 108) $= 20 \times 27$

LCM(90, 108) $= 540$

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The LCM of 90 and 108 is 540.

We will find the HCF of 162, 108, and 54 using the prime factorisation method, and their LCM using the common division method.

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Finding HCF using Prime Factorisation:

Step 1: Find the prime factorisation of each number.

Prime factorisation of 162:

$\begin{array}{c|cc} 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$162 = 2^1 \times 3^4$

Prime factorisation of 108 (from a previous example):

$108 = 2^2 \times 3^3$

Prime factorisation of 54 (from a previous example):

$54 = 2^1 \times 3^3$

---

Step 2: Identify common prime factors.

The prime factors common to 162, 108, and 54 are 2 and 3.

---

Step 3: Take the lowest power of each common prime factor.

For prime factor 2: Powers are $2^1$ (in 162), $2^2$ (in 108), $2^1$ (in 54). Lowest power is $2^1$.

For prime factor 3: Powers are $3^4$ (in 162), $3^3$ (in 108), $3^3$ (in 54). Lowest power is $3^3$.

---

Step 4: Multiply the lowest powers.

HCF(162, 108, 54) $= 2^1 \times 3^3 = 2 \times (3 \times 3 \times 3) = 2 \times 27 = 54$

The HCF of 162, 108, and 54 is 54.

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Finding LCM using Common Division Method:

We arrange 162, 108, and 54 and divide by prime factors:

$\begin{array}{c|ccc} 2 & 162 \;, & 108 \;, & 54 \\ \hline 3 & 81 \; , & 54 \; , & 27 \\ \hline 3 & 27 \; , & 18 \; , & 9 \\ \hline 3 & 9 \; , & 6 \; , & 3 \\ \hline 3 & 3 \; , & 2 \; , & 1 \\ \hline 2 & 1 \; , & 2 \; , & 1 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

---

The prime divisors used are 2, 3, 3, 3, 3, and 2.

To find the LCM, we multiply all these divisors:

LCM(162, 108, 54) $= 2 \times 3 \times 3 \times 3 \times 3 \times 2$

LCM(162, 108, 54) $= (2 \times 2) \times (3 \times 3 \times 3 \times 3)$

LCM(162, 108, 54) $= 4 \times 81$

LCM(162, 108, 54) $= 324$

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The LCM of 162, 108, and 54 is 324.

Relation between HCF and LCM of Two Numbers:

For any two positive integers, the product of the numbers is equal to the product of their Highest Common Factor (HCF) and Least Common Multiple (LCM).

If the two numbers are $a$ and $b$, then the relation is:

$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$

---

Finding HCF and LCM of 32 and 56:

We will use the prime factorisation method.

Prime factorisation of 32:

$\begin{array}{c|cc} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

$32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$

Prime factorisation of 56:

$\begin{array}{c|cc} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7^1$

---

To find the HCF, take the lowest power of the common prime factors.

The only common prime factor is 2. The lowest power is $2^3$.

HCF(32, 56) $= 2^3 = 8$

---

To find the LCM, take the highest power of all prime factors involved.

Prime factors involved are 2 and 7. Highest power of 2 is $2^5$. Highest power of 7 is $7^1$.

LCM(32, 56) $= 2^5 \times 7^1 = 32 \times 7 = 224$

---

HCF(32, 56) = 8

LCM(32, 56) = 224

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Verification of the Relation:

Product of the two numbers $= 32 \times 56$

$\begin{array}{cc}& & 3 & 2 \\ \times & & 5 & 6 \\ \hline && 1 & 9 & 2 \\ 1 & 6 & 0 & \times \\ \hline 1 & 7 & 9 & 2 \\ \hline \end{array}$

Product of the two numbers = 1792

---

Product of HCF and LCM $= \text{HCF}(32, 56) \times \text{LCM}(32, 56)$

Product of HCF and LCM $= 8 \times 224$

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

Product of HCF and LCM = 1792

---

Since Product of the two numbers ($1792$) = Product of HCF and LCM ($1792$), the relation is verified for the numbers 32 and 56.

To find the maximum capacity of a container that can measure the kerosene oil from both tankers an exact number of times, we need to find the Highest Common Factor (HCF) of the capacities of the two tankers (850 litres and 680 litres).

---

We will use the prime factorisation method to find the HCF.

Step 1: Prime factorisation of 850

$\begin{array}{c|cc} 2 & 850 \\ \hline 5 & 425 \\ \hline 5 & 85 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$

So, $850 = 2 \times 5 \times 5 \times 17 = 2^1 \times 5^2 \times 17^1$

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Step 2: Prime factorisation of 680

$\begin{array}{c|cc} 2 & 680 \\ \hline 2 & 340 \\ \hline 2 & 170 \\ \hline 5 & 85 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$

So, $680 = 2 \times 2 \times 2 \times 5 \times 17 = 2^3 \times 5^1 \times 17^1$

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Step 3: Identify common prime factors and take the lowest power

Common prime factors are 2, 5, and 17.

Lowest power of 2: $2^1$ (from 850)

Lowest power of 5: $5^1$ (from 680)

Lowest power of 17: $17^1$ (from both)

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Step 4: Calculate the HCF

HCF(850, 680) $= 2^1 \times 5^1 \times 17^1 = 2 \times 5 \times 17 = 10 \times 17 = 170$

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The maximum capacity of the container that can measure the kerosene oil of both tankers exactly is 170 litres.

We are looking for the least number which, when divided by 12, 16, and 24, leaves a remainder of 5 in each case.

Let the required number be $N$.

According to the problem, when $N$ is divided by 12, the remainder is 5. This means $N - 5$ is divisible by 12.

Similarly, when $N$ is divided by 16, the remainder is 5. This means $N - 5$ is divisible by 16.

And when $N$ is divided by 24, the remainder is 5. This means $N - 5$ is divisible by 24.

So, $N - 5$ is a number that is divisible by 12, 16, and 24. The least such number is the Least Common Multiple (LCM) of 12, 16, and 24.

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We need to find the LCM of 12, 16, and 24 using the common division method.

$\begin{array}{c|ccc} 2 & 12 \;, & 16 \;, & 24 \\ \hline 2 & 6 \; , & 8 \; , & 12 \\ \hline 2 & 3 \; , & 4 \; , & 6 \\ \hline 2 & 3 \; , & 2 \; , & 3 \\ \hline 3 & 3 \; , & 1 \; , & 3 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

The prime divisors used are 2, 2, 2, 2, and 3.

LCM(12, 16, 24) $= 2 \times 2 \times 2 \times 2 \times 3 = 16 \times 3 = 48$

So, the least number that is divisible by 12, 16, and 24 is 48.

This means $N - 5 = 48$.

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To find $N$, we add 5 to 48:

$N = 48 + 5 = 53$

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The least number which when divided by 12, 16, and 24 leaves a remainder 5 in each case is 53.

Check:

$53 \div 12 = 4$ with remainder $53 - (12 \times 4) = 53 - 48 = 5$.

$53 \div 16 = 3$ with remainder $53 - (16 \times 3) = 53 - 48 = 5$.

$53 \div 24 = 2$ with remainder $53 - (24 \times 2) = 53 - 48 = 5$.

To find the longest tape that can measure the three dimensions of the room exactly, we need to find the Highest Common Factor (HCF) of the length, breadth, and height of the room.

The dimensions are given as 600 cm, 450 cm, and 300 cm.

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We will find the HCF of 600, 450, and 300 using the prime factorisation method.

Step 1: Prime factorisation of 600

$\begin{array}{c|cc} 2 & 600 \\ \hline 2 & 300 \\ \hline 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$600 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 = 2^3 \times 3^1 \times 5^2$

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Step 2: Prime factorisation of 450

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

$450 = 2 \times 3 \times 3 \times 5 \times 5 = 2^1 \times 3^2 \times 5^2$

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Step 3: Prime factorisation of 300

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

$300 = 2 \times 2 \times 3 \times 5 \times 5 = 2^2 \times 3^1 \times 5^2$

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Step 4: Identify common prime factors and take the lowest power

The prime factors common to 600, 450, and 300 are 2, 3, and 5.

Lowest power of 2: $2^1$ (from 450)

Lowest power of 3: $3^1$ (from 600 and 300)

Lowest power of 5: $5^2$ (from 600, 450, and 300)

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Step 5: Calculate the HCF

HCF(600, 450, 300) $= 2^1 \times 3^1 \times 5^2 = 2 \times 3 \times 25 = 6 \times 25 = 150$

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The maximum capacity of the tape is 150 cm.

The longest tape which can measure the three dimensions of the room exactly is 150 cm.