Chapter 1 Real Numbers

The given rational number is $\frac{33}{2^2 \cdot 5}$.

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A rational number in the form $\frac{p}{q}$, where $p$ and $q$ are co-prime, has a terminating decimal expansion if the prime factorisation of $q$ is of the form $2^m \cdot 5^n$, where $m$ and $n$ are non-negative integers.

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In the given number, the denominator is $2^2 \cdot 5^1$. This is already in the form $2^m \cdot 5^n$ with $m=2$ and $n=1$.

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To determine the number of decimal places after which the expansion terminates, we need to make the powers of 2 and 5 in the denominator equal. The highest power is 2 (from $2^2$). We need to make the power of 5 equal to 2. We multiply the numerator and denominator by $5^{2-1} = 5^1 = 5$.

$\frac{33}{2^2 \cdot 5} = \frac{33 \times 5}{2^2 \cdot 5 \times 5} = \frac{165}{2^2 \cdot 5^2}$

$= \frac{165}{(2 \cdot 5)^2} = \frac{165}{10^2} = \frac{165}{100}$

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The decimal expansion is $\frac{165}{100} = 1.65$.

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The decimal expansion terminates after 2 decimal places.

Alternatively, the number of decimal places after which a rational number $\frac{p}{2^m 5^n}$ terminates is the maximum of $m$ and $n$. In this case, $m=2$ and $n=1$. The maximum of 2 and 1 is 2.

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Therefore, the decimal expansion terminates after two decimal places.

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The correct option is (B) two decimal places.

Euclid's division lemma states that for any two positive integers $a$ (dividend) and $b$ (divisor), there exist unique integers $q$ (quotient) and $r$ (remainder) satisfying the relation:

$a = bq + r$

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The condition on the remainder $r$ according to Euclid's division lemma is that $r$ must be greater than or equal to 0 and strictly less than the divisor $b$.

This condition is written as:

$0 \leq r < b$

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Comparing this condition with the given options:

(A) $1 < r < b$ - Incorrect, as $r$ can be 0 or 1.

(B) $0 < r \leq b$ - Incorrect, as $r$ can be 0 and $r$ must be strictly less than $b$.

(C) $0 \leq r < b$ - Correct, this matches the condition from Euclid's division lemma.

(D) $0 < r < b$ - Incorrect, as $r$ can be equal to 0.

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Therefore, the remainder $r$ must satisfy the condition $0 \leq r < b$.

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The correct option is (C) $0 \leq r < b$.

An integer is defined as even if it is divisible by 2. This means an even integer can be written as 2 multiplied by some integer.

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Let $m$ be any integer. We need to determine which of the given forms always represents an even integer.

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Let's examine each option:

(A) $m$: This form can represent either an even or an odd integer, depending on the value of $m$. For example, if $m=3$, it is odd; if $m=4$, it is even.

(B) $m + 1$: If $m$ is even, $m+1$ is odd. If $m$ is odd, $m+1$ is even. This form does not always represent an even integer.

(C) $2m$: For any integer value of $m$, the product $2m$ is always divisible by 2. This is the definition of an even integer. For example, if $m=-1$, $2m=-2$ (even); if $m=0$, $2m=0$ (even); if $m=5$, $2m=10$ (even).

(D) $2m + 1$: For any integer value of $m$, $2m$ is even, so $2m+1$ is always odd. This is the definition of an odd integer.

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Therefore, for some integer $m$, every even integer is of the form $2m$.

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The correct option is (C) 2m.

An integer is defined as odd if it is not divisible by 2. This means an odd integer leaves a remainder of 1 when divided by 2.

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Let $q$ be any integer. We need to determine which of the given forms always represents an odd integer.

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Let's examine each option:

(A) $q$: This form can represent either an odd or an even integer, depending on the value of $q$. For example, if $q=5$, it is odd; if $q=6$, it is even.

(B) $q + 1$: If $q$ is odd, $q+1$ is even. If $q$ is even, $q+1$ is odd. This form does not always represent an odd integer.

(C) $2q$: For any integer value of $q$, the product $2q$ is always divisible by 2, meaning it is always even. For example, if $q=-2$, $2q=-4$ (even); if $q=0$, $2q=0$ (even); if $q=3$, $2q=6$ (even).

(D) $2q + 1$: For any integer value of $q$, $2q$ is even. Adding 1 to an even number always results in an odd number. For example, if $q=-1$, $2q+1=-2+1=-1$ (odd); if $q=0$, $2q+1=0+1=1$ (odd); if $q=4$, $2q+1=8+1=9$ (odd).

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Therefore, for some integer $q$, every odd integer is of the form $2q + 1$.

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The correct option is (D) 2q + 1.

We are asked for the condition on the integer $n$ such that $n^2 - 1$ is divisible by 8.

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For $n^2 - 1$ to be divisible by 8, it must be equal to $8k$ for some integer $k$.

$n^2 - 1 = 8k$

$n^2 = 8k + 1$

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This implies that $n^2$ must be an odd number, since it is of the form (an even number) + 1. The square of an integer is odd if and only if the integer itself is odd.

Therefore, $n$ must be an odd integer.

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Let's verify this by considering the case when $n$ is an odd integer.

Any odd integer $n$ can be written in the form $n = 2k + 1$ for some integer $k$.

Substitute this into the expression $n^2 - 1$:

$n^2 - 1 = (2k + 1)^2 - 1$

$= ( (2k)^2 + 2(2k)(1) + 1^2 ) - 1$

$= (4k^2 + 4k + 1) - 1$

$= 4k^2 + 4k$

$= 4k(k + 1)$

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Now, consider the term $k(k+1)$. This is the product of two consecutive integers. The product of any two consecutive integers is always an even number, because one of the integers ($k$ or $k+1$) must be even.

So, $k(k+1) = 2m$ for some integer $m$.

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Substitute this back into the expression for $n^2 - 1$:

$n^2 - 1 = 4 \times (2m)$

$= 8m$

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Since $n^2 - 1$ can be written in the form $8m$, where $m$ is an integer, it is divisible by 8 when $n$ is an odd integer.

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Let's consider the case when $n$ is an even integer to confirm. If $n$ is even, $n = 2k$ for some integer $k$.

$n^2 - 1 = (2k)^2 - 1 = 4k^2 - 1$.

If $k=1$, $n=2$, $4(1)^2 - 1 = 3$, not divisible by 8.

If $k=2$, $n=4$, $4(2)^2 - 1 = 15$, not divisible by 8.

If $k=3$, $n=6$, $4(3)^2 - 1 = 35$, not divisible by 8.

For any integer $k$, $4k^2$ is a multiple of 4. $4k^2 - 1$ will always leave a remainder of $3$ when divided by 4, and thus cannot be divisible by 8.

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Therefore, $n^2 - 1$ is divisible by 8 if and only if $n$ is an odd integer.

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The correct option is (C) an odd integer.

First, we need to find the HCF (Highest Common Factor) of 65 and 117.

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We can use the prime factorisation method or Euclid's division algorithm to find the HCF.

Using Euclid's division algorithm:

Divide 117 by 65:

$117 = 65 \times 1 + 52$

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Now, divide the divisor 65 by the remainder 52:

$65 = 52 \times 1 + 13$

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Now, divide the divisor 52 by the remainder 13:

$52 = 13 \times 4 + 0$

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Since the remainder is 0, the HCF is the last non-zero remainder, which is 13.

So, HCF(65, 117) = 13.

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Alternatively, using prime factorisation:

Prime factors of 65: $65 = 5 \times 13$

Prime factors of 117: $117 = 3 \times 39 = 3 \times 3 \times 13 = 3^2 \times 13$

The common prime factor is 13 with the lowest power being $13^1$.

So, HCF(65, 117) = 13.

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We are given that the HCF is expressible in the form $65m - 117$.

Equating the HCF we found with the given form:

$65m - 117 = 13$

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Now, we solve for $m$:

Add 117 to both sides of the equation:

$65m = 13 + 117$

$65m = 130$

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Divide both sides by 65:

$m = \frac{130}{65}$

$m = 2$

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The value of $m$ is 2.

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The correct option is (B) 2.

Let the largest number be $N$.

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According to the problem statement, when 70 is divided by $N$, the remainder is 5. This implies that $70 - 5$ is perfectly divisible by $N$.

$70 - 5 = 65$

So, $N$ is a divisor of 65.

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Similarly, when 125 is divided by $N$, the remainder is 8. This implies that $125 - 8$ is perfectly divisible by $N$.

$125 - 8 = 117$

So, $N$ is a divisor of 117.

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Since $N$ is a divisor of both 65 and 117, it is a common divisor of 65 and 117.

We are looking for the largest such number, which means we need to find the Highest Common Factor (HCF) of 65 and 117.

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We can find the HCF of 65 and 117 using prime factorisation:

Prime factors of 65: $65 = 5 \times 13$

Prime factors of 117: $117 = 3 \times 39 = 3 \times 3 \times 13 = 3^2 \times 13$

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The common prime factor is 13, with the lowest power being $13^1$.

Therefore, HCF(65, 117) = 13.

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The largest number that divides 65 and 117 is 13. This number, 13, is the required number $N$.

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We must also check if this number $N=13$ is greater than the remainders (5 and 8). Since $13 > 5$ and $13 > 8$, the condition is satisfied.

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Thus, the largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is 13.

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The correct option is (A) 13.

We are given two positive integers $a$ and $b$ in terms of their prime factorisation, where $x$ and $y$ are prime numbers:

$a = x^3 y^2$

$b = x y^3$

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To find the Highest Common Factor (HCF) of two numbers given by their prime factorisations, we take the product of the common prime factors, each raised to the lowest power that appears in the factorisations of the numbers.

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The prime factors present in both $a$ and $b$ are $x$ and $y$.

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Let's look at the powers of each common prime factor:

For the prime factor $x$:
In the factorisation of $a$, the power of $x$ is 3 ($x^3$).
In the factorisation of $b$, the power of $x$ is 1 ($x^1$).
The lowest power of $x$ is $\min(3, 1) = 1$. So we take $x^1$.

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For the prime factor $y$:
In the factorisation of $a$, the power of $y$ is 2 ($y^2$).
In the factorisation of $b$, the power of $y$ is 3 ($y^3$).
The lowest power of $y$ is $\min(2, 3) = 2$. So we take $y^2$.

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The HCF(a, b) is the product of these terms with the lowest powers.

HCF(a, b) = $x^1 \times y^2 = xy^2$

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Thus, the HCF of $a$ and $b$ is $xy^2$.

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The correct option is (B) $xy^2$.

We are given two positive integers $p$ and $q$ in terms of their prime factorisation, where $a$ and $b$ are prime numbers:

$p = ab^2 = a^1 b^2$

$q = a^3b = a^3 b^1$

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To find the Least Common Multiple (LCM) of two numbers given by their prime factorisations, we take the product of all distinct prime factors present in either factorisation, each raised to the highest power that appears in the factorisations of the numbers.

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The distinct prime factors present in the factorisations of $p$ and $q$ are $a$ and $b$.

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Let's determine the highest power for each distinct prime factor:

For the prime factor $a$:
In the factorisation of $p$, the power of $a$ is 1 ($a^1$).
In the factorisation of $q$, the power of $a$ is 3 ($a^3$).
The highest power of $a$ is $\max(1, 3) = 3$. So we take $a^3$.

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For the prime factor $b$:
In the factorisation of $p$, the power of $b$ is 2 ($b^2$).
In the factorisation of $q$, the power of $b$ is 1 ($b^1$).
The highest power of $b$ is $\max(2, 1) = 2$. So we take $b^2$.

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The LCM(p, q) is the product of these terms with the highest powers.

LCM(p, q) = $a^3 \times b^2 = a^3b^2$

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Thus, the LCM of $p$ and $q$ is $a^3b^2$.

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The correct option is (C) a³b².

Let $r$ be a non-zero rational number and $i$ be an irrational number.

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By definition, a rational number can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Since $r$ is non-zero, $p$ must also be non-zero.

So, $r = \frac{a}{b}$ for some integers $a$ and $b$, where $a \neq 0$ and $b \neq 0$.

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An irrational number cannot be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

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We want to determine the nature of the product $r \cdot i$.

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Let's assume, for the sake of contradiction, that the product $r \cdot i$ is a rational number.

If $r \cdot i$ is rational, it can be written as $\frac{c}{d}$ for some integers $c$ and $d$, where $d \neq 0$.

So, we have:

$r \cdot i = \frac{c}{d}$

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Substitute $r = \frac{a}{b}$ into the equation:

$\frac{a}{b} \cdot i = \frac{c}{d}$

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Since $r$ is non-zero, $a \neq 0$. We can multiply both sides by $\frac{b}{a}$ (which is defined since $a \neq 0$ and $b \neq 0$) to isolate $i$:

$i = \frac{c}{d} \cdot \frac{b}{a}$

$i = \frac{cb}{da}$

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Since $c, b, d, a$ are integers, and $d \neq 0$, $a \neq 0$, the product $da$ is also an integer and $da \neq 0$.

The expression $\frac{cb}{da}$ represents a ratio of two integers ($cb$ and $da$) with a non-zero denominator. This means that $i$ is a rational number.

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This conclusion contradicts our initial statement that $i$ is an irrational number.

Therefore, our initial assumption that the product $r \cdot i$ is rational must be false.

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Hence, the product of a non-zero rational number and an irrational number must always be an irrational number.

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The correct option is (A) always irrational.

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is the Least Common Multiple (LCM) of these numbers.

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We need to find the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

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To find the LCM, we first find the prime factorisation of each number:

$1 = 1$

$2 = 2^1$

$3 = 3^1$

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

$5 = 5^1$

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

$7 = 7^1$

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

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

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

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

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Now, we find the highest power of each prime factor that appears in any of the factorisations:

Highest power of 2: From 2, 4, 6, 8, 10, the powers are $2^1, 2^2, 2^1, 2^3, 2^1$. The highest power is $2^3 = 8$.

Highest power of 3: From 3, 6, 9, the powers are $3^1, 3^1, 3^2$. The highest power is $3^2 = 9$.

Highest power of 5: From 5, 10, the powers are $5^1, 5^1$. The highest power is $5^1 = 5$.

Highest power of 7: From 7, the power is $7^1$. The highest power is $7^1 = 7$.

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The LCM is the product of these highest powers:

LCM(1, 2, ..., 10) $= 2^3 \times 3^2 \times 5^1 \times 7^1$

$= 8 \times 9 \times 5 \times 7$

$= (8 \times 9) \times (5 \times 7)$

$= 72 \times 35$

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Calculating the product:

$72 \times 35 = 2520$

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Thus, the least number that is divisible by all the numbers from 1 to 10 is 2520.

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The correct option is (D) 2520.

The given rational number is $\frac{14587}{1250}$.

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A rational number has a terminating decimal expansion if and only if its denominator, in its simplest form, has prime factors only 2 and 5.

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First, let's find the prime factorisation of the denominator, 1250:

$1250 = 10 \times 125 = (2 \times 5) \times 5^3 = 2^1 \times 5^4$

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The denominator is $1250 = 2^1 \cdot 5^4$. The prime factors are only 2 and 5. Thus, the decimal expansion is terminating.

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To determine the number of decimal places after which the expansion terminates, we look at the powers of 2 and 5 in the denominator. The number of decimal places is the maximum of these powers.

In the denominator $2^1 \cdot 5^4$, the power of 2 is $m=1$ and the power of 5 is $n=4$.

The number of decimal places is $\max(m, n) = \max(1, 4) = 4$.

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We should also ensure the fraction is in simplest form. The denominator's prime factors are 2 and 5. The numerator 14587 is not divisible by 2 (it's odd) and not divisible by 5 (it doesn't end in 0 or 5). Therefore, there are no common prime factors between the numerator and the denominator, and the fraction is already in its simplest form.

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The decimal expansion of $\frac{14587}{1250}$ will terminate after a number of decimal places equal to the highest power of 2 or 5 in the denominator's prime factorisation, which is 4.

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Therefore, the decimal expansion will terminate after four decimal places.

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The correct option is (D) four decimal places.

According to Euclid’s division lemma, for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where the remainder $r$ satisfies the condition $0 \leq r < b$.

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In this problem, a positive integer $a$ is divided by 3. This means the divisor $b = 3$.

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Applying Euclid's division lemma with $b=3$, we have $a = 3q + r$, where the remainder $r$ must satisfy:

$0 \leq r < 3$

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The integers that satisfy this condition are 0, 1, and 2.

So, the possible values for the remainder $r$ when a positive integer $a$ is divided by 3 are 0, 1, and 2.

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For example:

If $a=3$, $3 = 3 \times 1 + 0$. Remainder is 0.

If $a=4$, $4 = 3 \times 1 + 1$. Remainder is 1.

If $a=5$, $5 = 3 \times 1 + 2$. Remainder is 2.

If $a=6$, $6 = 3 \times 2 + 0$. Remainder is 0.

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The statement says that the values of the remainder $r$ are 0 and 1 only.

This statement is false.

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Justification: By Euclid's division lemma, when a positive integer $a$ is divided by 3, the remainder $r$ must satisfy $0 \leq r < 3$. This inequality means that the possible integer values for $r$ are 0, 1, and 2. The set of possible remainders is $\{0, 1, 2\}$, not $\{0, 1\}$.

For any natural number $n$, the number $6^n$ is formed by multiplying 6 by itself $n$ times.

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A number ends with the digit 5 if and only if it is divisible by 5.

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For a number to be divisible by 5, its prime factorisation must contain the prime number 5.

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Let's find the prime factorisation of the base of the given number, which is 6:

$6 = 2 \times 3$

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Now, let's find the prime factorisation of $6^n$:

$6^n = (2 \times 3)^n = 2^n \times 3^n$

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The prime factors of $6^n$ are only 2 and 3, regardless of the value of the natural number $n$. The prime factor 5 is not present in the prime factorisation of $6^n$.

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By the Fundamental Theorem of Arithmetic, every composite number has a unique prime factorisation (apart from the order of the factors). Since the prime factorisation of $6^n$ does not contain 5, $6^n$ is not divisible by 5.

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Since $6^n$ is not divisible by 5, it cannot end with the digit 5.

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Therefore, the number $6^n$, for any natural number $n$, cannot end with the digit 5.

The statement "every positive integer can be of the form $4q + 2$, where $q$ is an integer" is false.

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According to Euclid’s division lemma, for any positive integer $a$ and a divisor $b=4$, there exist unique integers $q$ and $r$ such that $a = 4q + r$, where the remainder $r$ must satisfy the condition $0 \leq r < 4$.

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The possible integer values for the remainder $r$ in this case are 0, 1, 2, and 3.

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Therefore, any positive integer $a$ can be expressed in one of the following four forms:

$4q + 0 = 4q$

$4q + 1$

$4q + 2$

$4q + 3$

for some integer $q \geq 0$ (since $a$ is positive).

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The statement claims that every positive integer is *only* of the form $4q + 2$. This is incorrect because positive integers can also be of the forms $4q$, $4q + 1$, and $4q + 3$.

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For example:

The positive integer 1 is of the form $4q+1$ (when $q=0$, $4(0)+1=1$). It cannot be written in the form $4q+2$ for any integer $q$.

The positive integer 4 is of the form $4q$ (when $q=1$, $4(1)=4$). It cannot be written in the form $4q+2$ for any integer $q$.

The positive integer 7 is of the form $4q+3$ (when $q=1$, $4(1)+3=7$). It cannot be written in the form $4q+2$ for any integer $q$.

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Thus, not every positive integer can be of the form $4q + 2$. Only those positive integers that leave a remainder of 2 when divided by 4 are of this form.

The statement “The product of two consecutive positive integers is divisible by 2” is True.

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Reason:

Consider any two consecutive positive integers. Let these integers be $n$ and $n+1$, where $n$ is a positive integer.

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We know that among any two consecutive integers, one integer must be even and the other integer must be odd.

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Let's consider the two possible cases:

Case 1: The first integer $n$ is even.

If $n$ is even, then $n$ can be expressed in the form $n = 2k$ for some positive integer $k$.

The product of the two consecutive integers is $n(n+1)$. Substituting $n=2k$, we get:

$n(n+1) = (2k)(n+1)$

Since the expression $(2k)(n+1)$ has a factor of 2, the product $n(n+1)$ is divisible by 2.

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Case 2: The first integer $n$ is odd.

If $n$ is odd, then the next consecutive integer, $n+1$, must be even.

If $n+1$ is even, then $n+1$ can be expressed in the form $n+1 = 2k$ for some positive integer $k$.

The product of the two consecutive integers is $n(n+1)$. Substituting $n+1=2k$, we get:

$n(n+1) = n(2k)$

Since the expression $n(2k)$ has a factor of 2, the product $n(n+1)$ is divisible by 2.

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In both cases, the product of two consecutive positive integers is always divisible by 2.

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Therefore, the statement is True.

The statement “The product of three consecutive positive integers is divisible by 6” is True.

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Reason:

Let the three consecutive positive integers be $n$, $n+1$, and $n+2$, where $n$ is a positive integer.

Their product is $P = n(n+1)(n+2)$.

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For a number to be divisible by 6, it must be divisible by both 2 and 3, since HCF(2, 3) = 1 and $6 = 2 \times 3$.

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Let's examine the divisibility of the product by 2 and 3 separately.

Divisibility by 2:

Among any two consecutive integers, one must be even and the other must be odd. In the set $\{n, n+1, n+2\}$, we have at least one pair of consecutive integers ($n, n+1$) or ($n+1, n+2$).

Alternatively, consider $n$. If $n$ is even, the product $n(n+1)(n+2)$ is even. If $n$ is odd, then $n+1$ is even, and the product $n(n+1)(n+2)$ is even. In any case, at least one of the three consecutive integers is even.

Since at least one factor in the product $n(n+1)(n+2)$ is even, the product is always divisible by 2.

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Divisibility by 3:

Among any three consecutive integers, exactly one must be a multiple of 3.

We can show this using Euclid's division lemma. When any integer $n$ is divided by 3, the remainder $r$ can be 0, 1, or 2. Thus, $n$ can be of the form $3k$, $3k+1$, or $3k+2$ for some integer $k$.

• If $n = 3k$, then $n$ is divisible by 3. The product $n(n+1)(n+2)$ is divisible by 3.
• If $n = 3k+1$, then $n+2 = (3k+1) + 2 = 3k+3 = 3(k+1)$. Thus, $n+2$ is divisible by 3. The product $n(n+1)(n+2)$ is divisible by 3.
• If $n = 3k+2$, then $n+1 = (3k+2) + 1 = 3k+3 = 3(k+1)$. Thus, $n+1$ is divisible by 3. The product $n(n+1)(n+2)$ is divisible by 3.

In all possible cases, at least one of the three consecutive integers is divisible by 3. Thus, the product $n(n+1)(n+2)$ is always divisible by 3.

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Since the product of three consecutive positive integers is divisible by both 2 and 3, it is also divisible by their product, which is $2 \times 3 = 6$.

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Therefore, the statement is True.

The statement "the square of any positive integer can be of the form $3m + 2$, where $m$ is a natural number" is false.

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Justification:

Let $n$ be any positive integer. According to Euclid’s division lemma, when a positive integer $n$ is divided by 3, the remainder can be 0, 1, or 2. Thus, any positive integer $n$ can be written in one of the following three forms, for some integer $k \geq 0$:

• $n = 3k$
• $n = 3k + 1$
• $n = 3k + 2$

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Let's find the square of $n$ in each of these cases:

Case 1: $n = 3k$

$n^2 = (3k)^2 = 9k^2 = 3(3k^2)$.

Let $m = 3k^2$. Since $k \geq 0$, $m$ is an integer $\geq 0$. If $n$ is a positive integer, then $k$ must be a positive integer ($k \geq 1$). In this case, $m = 3k^2$ is a natural number. The form is $3m$.

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Case 2: $n = 3k + 1$

$n^2 = (3k + 1)^2 = (3k)^2 + 2(3k)(1) + 1^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$.

Let $m = 3k^2 + 2k$. Since $k \geq 0$, $m$ is an integer $\geq 0$. If $n$ is a positive integer, $k \geq 0$. If $k=0$, $n=1$, $n^2=1 = 3(0)+1$, $m=0$ which is not a natural number. If $k \geq 1$, $m = 3k^2 + 2k$ is a natural number. The form is $3m + 1$.

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Case 3: $n = 3k + 2$

$n^2 = (3k + 2)^2 = (3k)^2 + 2(3k)(2) + 2^2 = 9k^2 + 12k + 4 = 9k^2 + 12k + 3 + 1 = 3(3k^2 + 4k + 1) + 1$.

Let $m = 3k^2 + 4k + 1$. Since $k \geq 0$, $3k^2 \geq 0$, $4k \geq 0$, and 1 is positive, $m \geq 1$. Thus, $m$ is always a natural number in this case for positive $n$. The form is $3m + 1.

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From the three cases, we see that the square of any positive integer $n$ is either of the form $3m$ or $3m + 1$ for some integer $m \geq 0$. Specifically, if $n$ is a multiple of 3, $n^2$ is a multiple of 3 (form $3m$ with $m \geq 1$). If $n$ is not a multiple of 3, $n^2$ leaves a remainder of 1 when divided by 3 (form $3m+1$ with $m \geq 0$, and $m \geq 1$ if $n>1$).

The square of a positive integer can never be of the form $3m + 2$, which corresponds to leaving a remainder of 2 when divided by 3.

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The condition that $m$ is a natural number ($m \geq 1$) doesn't allow for the case $n^2=1$ (where $m=0$). However, even with this restriction on $m$, the possible forms of $n^2$ when divided by 3 still only result in remainders 0 or 1, never 2. For example, $1^2=1 = 3m+1$ (m=0, not natural), $2^2=4=3(1)+1$ (m=1, natural), $3^2=9=3(3)$ (m=3, natural), $4^2=16=3(5)+1$ (m=5, natural), $5^2=25=3(8)+1$ (m=8, natural), $6^2=36=3(12)$ (m=12, natural).

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Thus, the square of any positive integer cannot leave a remainder of 2 when divided by 3, and therefore cannot be of the form $3m + 2$ for any integer $m$, including natural numbers.

No, the square of a positive integer of the form $3q + 1$, where $q$ is a natural number, cannot be written in any form other than $3m + 1$ for some integer $m$.

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Justification:

Let the positive integer be $n$. We are given that $n$ is of the form $3q + 1$, where $q$ is a natural number. A natural number is a positive integer, so $q \in \{1, 2, 3, \dots\}$.

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We want to find the form of the square of this integer, $n^2$.

$n = 3q + 1$

Squaring both sides:

$n^2 = (3q + 1)^2$

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Expand the expression using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$n^2 = (3q)^2 + 2(3q)(1) + 1^2$

$n^2 = 9q^2 + 6q + 1$

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We can factor out 3 from the first two terms:

$n^2 = 3(3q^2 + 2q) + 1$

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Let $m = 3q^2 + 2q$. Since $q$ is a natural number, $q$ is an integer. The term $3q^2$ is an integer, and the term $2q$ is an integer. The sum of two integers is always an integer. Therefore, $m = 3q^2 + 2q$ is an integer.

---

Substituting $m$ back into the expression for $n^2$, we get:

$n^2 = 3m + 1$

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This shows that the square of any positive integer of the form $3q + 1$ (where $q$ is a natural number) is always of the form $3m + 1$, where $m$ is the integer $3q^2 + 2q$.

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A number of the form $3m+1$ leaves a remainder of 1 when divided by 3.

A number of the form $3m$ leaves a remainder of 0 when divided by 3.

A number of the form $3m+2$ leaves a remainder of 2 when divided by 3.

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Since $n^2$ is always of the form $3m+1$, it will always leave a remainder of 1 when divided by 3. It can never leave a remainder of 0 or 2 when divided by 3.

---

Therefore, its square cannot be written in the form $3m$ or $3m + 2$ for any integer $m$.

We are given that the numbers 525 and 3000 are both divisible only by the numbers in the set $\{3, 5, 15, 25, 75\}$.

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This means that the set of common divisors of 525 and 3000 is precisely $\{3, 5, 15, 25, 75\}$.

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

In other words, the HCF is the largest number in the set of common divisors.

---

The given set of common divisors is $\{3, 5, 15, 25, 75\}$.

Let's identify the largest number in this set.

Comparing the numbers: 3, 5, 15, 25, 75.

The largest number in this set is 75.

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Therefore, the HCF of 525 and 3000 is 75.

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Justification:

By definition, the HCF is the largest common divisor. The problem explicitly states that the only common divisors of 525 and 3000 are 3, 5, 15, 25, and 75. Among these common divisors, 75 is the largest number. Hence, 75 is the HCF of 525 and 3000.

We can also verify this by checking the divisibility of 525 and 3000 by 75:

$525 \div 75 = 7$

$3000 \div 75 = 40$

Since both divisions result in integers (7 and 40), 75 is indeed a common divisor. And based on the information provided in the question, it is the largest among the common divisors.

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HCF (525, 3000) = 75.

A composite number is a natural number greater than 1 that is not prime. This means a composite number has at least one positive divisor other than 1 and itself.

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Let the given number be $N$.

$N = 3 \times 5 \times 7 + 7$

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We can factor out the common term '7' from both parts of the expression:

$N = (3 \times 5 \times 7) + (7 \times 1)$

$N = 7 \times (3 \times 5 + 1)$

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Now, simplify the expression inside the parentheses:

$3 \times 5 + 1 = 15 + 1 = 16$

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So, the number $N$ can be written as:

$N = 7 \times 16$

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The value of the number is $7 \times 16 = 112$.

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Since $N = 7 \times 16$, we can see that $N$ has factors 7 and 16. Both 7 and 16 are positive integers greater than 1 and less than 112. For example, 7 is a divisor of 112, and $112 \div 7 = 16$.

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Since the number $3 \times 5 \times 7 + 7$ (which equals 112) has positive divisors other than 1 and itself (specifically, 7 and 16, among others like 2, 4, 8, 14, 28, 56), it fits the definition of a composite number.

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Therefore, $3 \times 5 \times 7 + 7$ is a composite number because it can be expressed as a product of two integers greater than 1.

No, two numbers cannot have 18 as their HCF and 380 as their LCM.

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Reason:

For any two positive integers, it is a fundamental property that their HCF (Highest Common Factor) must always be a factor of their LCM (Least Common Multiple).

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In this case, the given HCF is 18 and the given LCM is 380.

We need to check if 18 divides 380 without leaving a remainder.

---

Let's perform the division:

Divide 380 by 18.

$\frac{380}{18} = \frac{190}{9}$

The fraction $\frac{190}{9}$ is not an integer because 190 is not divisible by 9 ($190 = 9 \times 21 + 1$).

Alternatively, we can check the divisibility of 380 by 18. A number is divisible by 18 if it is divisible by both 2 and 9.

380 is divisible by 2 (since it is an even number).

For divisibility by 9, the sum of the digits of 380 is $3 + 8 + 0 = 11$. Since 11 is not divisible by 9, 380 is not divisible by 9.

---

Since 380 is not divisible by 18, 18 is not a factor of 380.

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Because the given HCF (18) is not a factor of the given LCM (380), it is not possible for two numbers to have 18 as their HCF and 380 as their LCM.

To determine if a rational number has a terminating or non-terminating (repeating) decimal expansion without performing long division, we need to examine the prime factorisation of its denominator when the fraction is in its simplest form.

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A rational number $\frac{p}{q}$, where $p$ and $q$ are co-prime integers and $q \neq 0$, has a terminating decimal expansion if and only if the prime factorisation of the denominator $q$ is of the form $2^m \cdot 5^n$, where $m$ and $n$ are non-negative integers.

If the prime factorisation of $q$ contains any prime factor other than 2 or 5, the decimal expansion is non-terminating (repeating).

---

The given fraction is $\frac{987}{10500}$.

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First, we need to simplify the fraction by finding the HCF of the numerator (987) and the denominator (10500).

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Let's find the prime factorisation of the numerator 987:

Sum of digits of 987 is $9+8+7 = 24$, which is divisible by 3. So, 987 is divisible by 3.

$987 \div 3 = 329$

To factorise 329, we can test small prime numbers. 329 is not divisible by 2, 3 (sum of digits = 14), 5. Let's try 7.

$329 \div 7 = 47$

47 is a prime number.

So, the prime factorisation of 987 is $3 \times 7 \times 47$.

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Now, let's find the prime factorisation of the denominator 10500:

$10500 = 105 \times 100$

$105 = 3 \times 35 = 3 \times 5 \times 7$

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

So, $10500 = (3 \times 5 \times 7) \times (2^2 \times 5^2) = 2^2 \times 3^1 \times 5^3 \times 7^1$.

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Now, write the fraction with the prime factorisations:

$\frac{987}{10500} = \frac{3 \times 7 \times 47}{2^2 \times 3 \times 5^3 \times 7}$

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We can see that there are common factors of 3 and 7 in the numerator and the denominator. Let's cancel them out:

$\frac{\cancel{3} \times \cancel{7} \times 47}{2^2 \times \cancel{3} \times 5^3 \times \cancel{7}} = \frac{47}{2^2 \times 5^3}$

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The fraction in its simplest form is $\frac{47}{2^2 \times 5^3}$. The denominator of the simplest form is $q' = 2^2 \times 5^3$.

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The prime factors of the denominator $q'$ are only 2 and 5. This is in the form $2^m \cdot 5^n$ with $m=2$ and $n=3$.

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Therefore, the decimal expansion of the rational number $\frac{987}{10500}$ will be terminating.

The given decimal number is 327.7081.

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This is a terminating decimal expansion because the digits after the decimal point end after a finite number of steps (four digits in this case).

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We know that a rational number $\frac{p}{q}$ (where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are co-prime) has a terminating decimal expansion if and only if the prime factorisation of the denominator $q$ is of the form $2^m \cdot 5^n$, where $m$ and $n$ are non-negative integers.

---

Let's express the given decimal number as a fraction:

$327.7081 = \frac{3277081}{10000}$

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Now, we find the prime factorisation of the denominator, 10000:

$10000 = 10^4 = (2 \times 5)^4 = 2^4 \times 5^4$

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So, the denominator is $2^4 \times 5^4$. Its prime factors are only 2 and 5.

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To express the rational number in the form $\frac{p}{q}$ where $p$ and $q$ are co-prime, we need to check if the numerator 3277081 has any common factors with the denominator $2^4 \times 5^4$. The prime factors of the denominator are 2 and 5.

The numerator 3277081 ends in 1, so it is not divisible by 2.

The numerator 3277081 ends in 1, so it is not divisible by 5.

Since the numerator is not divisible by the only prime factors (2 and 5) of the denominator, the fraction $\frac{3277081}{10000}$ is already in its simplest form.

---

Therefore, when the number 327.7081 is expressed in the form $\frac{p}{q}$ where $p$ and $q$ are co-prime, the denominator $q$ is 10000, and its prime factorisation is $2^4 \times 5^4$.

---

Conclusion: The prime factors of $q$ are only 2 and 5.

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Reason: A terminating decimal expansion corresponds to a rational number whose denominator, in simplest form, has only prime factors 2 and/or 5.

Two positive integers are called co-prime (or relatively prime) if their Highest Common Factor (HCF) is 1.

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We will use Euclid's division algorithm to find the HCF for each pair of numbers. The algorithm states that for any two positive integers $a$ and $b$ with $a > b$, we can write $a = bq + r$, where $0 \leq r < b$. The HCF(a, b) = HCF(b, r). We repeat the process until the remainder is 0. The last non-zero remainder is the HCF.

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(i) HCF of 231 and 396

Let $a = 396$ and $b = 231$.

Step 1: Divide 396 by 231.

$396 = 231 \times 1 + 165$

The remainder is 165.

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Step 2: Divide 231 by 165.

$231 = 165 \times 1 + 66$

The remainder is 66.

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Step 3: Divide 165 by 66.

$165 = 66 \times 2 + 33$

The remainder is 33.

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Step 4: Divide 66 by 33.

$66 = 33 \times 2 + 0$

The remainder is 0.

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The last non-zero remainder is 33.

So, HCF(231, 396) = 33.

Since the HCF is 33 (which is not 1), the numbers 231 and 396 are not co-prime.

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(ii) HCF of 847 and 2160

Let $a = 2160$ and $b = 847$.

Step 1: Divide 2160 by 847.

$2160 = 847 \times 2 + 466$

The remainder is 466.

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Step 2: Divide 847 by 466.

$847 = 466 \times 1 + 381$

The remainder is 381.

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Step 3: Divide 466 by 381.

$466 = 381 \times 1 + 85$

The remainder is 85.

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Step 4: Divide 381 by 85.

$381 = 85 \times 4 + 41$

The remainder is 41.

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Step 5: Divide 85 by 41.

$85 = 41 \times 2 + 3$

The remainder is 3.

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Step 6: Divide 41 by 3.

$41 = 3 \times 13 + 2$

The remainder is 2.

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Step 7: Divide 3 by 2.

$3 = 2 \times 1 + 1$

The remainder is 1.

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Step 8: Divide 2 by 1.

$2 = 1 \times 2 + 0$

The remainder is 0.

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The last non-zero remainder is 1.

So, HCF(847, 2160) = 1.

Since the HCF is 1, the numbers 847 and 2160 are co-prime.

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Conclusion:

(i) The pair (231, 396) is not co-prime.

(ii) The pair (847, 2160) is co-prime.

Let $a$ be an odd positive integer.

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According to Euclid’s division lemma, any positive integer can be expressed in the form $bq + r$, where $0 \leq r < b$. Let us consider the divisor $b=4$. Any positive integer $a$ can be written in one of the forms $4q$, $4q+1$, $4q+2$, or $4q+3$, for some integer $q \geq 0$.

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Since $a$ is an odd positive integer, it cannot be of the form $4q$ (which is even) or $4q+2$ (which is also even, as $4q+2 = 2(2q+1)$).

Therefore, an odd positive integer $a$ must be of the form $4q + 1$ or $4q + 3$, for some integer $q \geq 0$.

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We need to show that the square of $a$ is of the form $8m + 1$ for some whole number $m$ (a whole number is a non-negative integer, i.e., $0, 1, 2, \dots$).

---

Case 1: $a = 4q + 1$ for some integer $q \geq 0$.

Consider the square of $a$:

$a^2 = (4q + 1)^2$

Expand the expression:

$a^2 = (4q)^2 + 2(4q)(1) + 1^2$

$a^2 = 16q^2 + 8q + 1$

Factor out 8 from the first two terms:

$a^2 = 8(2q^2 + q) + 1$

Let $m = 2q^2 + q$. Since $q$ is an integer and $q \geq 0$, $q^2 \geq 0$, $2q^2 \geq 0$, and $q \geq 0$. Therefore, $m = 2q^2 + q$ is a non-negative integer. Thus, $m$ is a whole number.

So, when $a = 4q+1$, $a^2$ is of the form $8m + 1$ for a whole number $m = 2q^2 + q$.

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Case 2: $a = 4q + 3$ for some integer $q \geq 0$.

Consider the square of $a$:

$a^2 = (4q + 3)^2$

Expand the expression:

$a^2 = (4q)^2 + 2(4q)(3) + 3^2$

$a^2 = 16q^2 + 24q + 9$

Rewrite the term 9 as $8+1$:

$a^2 = 16q^2 + 24q + 8 + 1$

Factor out 8 from the first three terms:

$a^2 = 8(2q^2 + 3q + 1) + 1$

Let $m = 2q^2 + 3q + 1$. Since $q$ is an integer and $q \geq 0$, $q^2 \geq 0$, $2q^2 \geq 0$, $3q \geq 0$, and 1 is a positive integer. Therefore, $m = 2q^2 + 3q + 1$ is a non-negative integer. Thus, $m$ is a whole number.

So, when $a = 4q+3$, $a^2$ is of the form $8m + 1$ for a whole number $m = 2q^2 + 3q + 1$.

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In both cases, the square of an odd positive integer is of the form $8m + 1$, where $m$ is a whole number.

This completes the proof.

To Prove: $\sqrt{2} + \sqrt{3}$ is irrational.

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Proof:

Assume, for the sake of contradiction, that $\sqrt{2} + \sqrt{3}$ is a rational number.

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If $\sqrt{2} + \sqrt{3}$ is rational, then we can write it in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and HCF$(p, q) = 1$ (i.e., the fraction is in its simplest form).

Let $\sqrt{2} + \sqrt{3} = \frac{p}{q}$

---

Rearrange the equation to isolate one of the radical terms. Let's isolate $\sqrt{2}$:

$\sqrt{2} = \frac{p}{q} - \sqrt{3}$

---

Square both sides of the equation to eliminate the radical on the left side:

$(\sqrt{2})^2 = \left(\frac{p}{q} - \sqrt{3}\right)^2$

$2 = \left(\frac{p}{q}\right)^2 - 2 \cdot \frac{p}{q} \cdot \sqrt{3} + (\sqrt{3})^2$

$2 = \frac{p^2}{q^2} - \frac{2p}{q} \sqrt{3} + 3$

---

Now, rearrange the equation to isolate the term containing $\sqrt{3}$:

$\frac{2p}{q} \sqrt{3} = \frac{p^2}{q^2} + 3 - 2$

$\frac{2p}{q} \sqrt{3} = \frac{p^2}{q^2} + 1$

Combine the terms on the right side:

$\frac{2p}{q} \sqrt{3} = \frac{p^2 + q^2}{q^2}$

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Now, solve for $\sqrt{3}$. Since $\frac{p}{q} = \sqrt{2} + \sqrt{3} \neq 0$, $p \neq 0$. Also $q \neq 0$. Thus, $2pq \neq 0$. We can multiply both sides by $\frac{q}{2p}$:

$\sqrt{3} = \frac{p^2 + q^2}{q^2} \times \frac{q}{2p}$

$\sqrt{3} = \frac{p^2 + q^2}{2pq}$

---

Consider the expression on the right side: $\frac{p^2 + q^2}{2pq}$.

Since $p$ and $q$ are integers, $p^2$ is an integer, $q^2$ is an integer, $p^2 + q^2$ is an integer. Similarly, $2pq$ is an integer.

Also, as noted, $2pq \neq 0$.

Therefore, the expression $\frac{p^2 + q^2}{2pq}$ is the ratio of two integers with a non-zero denominator. By definition, this expression represents a rational number.

---

So, the equation $\sqrt{3} = \frac{p^2 + q^2}{2pq}$ implies that an irrational number ($\sqrt{3}$) is equal to a rational number ($\frac{p^2 + q^2}{2pq}$).

This is a contradiction because an irrational number cannot be equal to a rational number. We know that $\sqrt{3}$ is irrational.

---

The contradiction arose from our initial assumption that $\sqrt{2} + \sqrt{3}$ is rational.

Therefore, the initial assumption must be false.

---

Hence, $\sqrt{2} + \sqrt{3}$ is an irrational number.

To Show: The square of any positive integer is either of the form $4q$ or $4q + 1$ for some integer $q$.

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Proof:

Let $a$ be any positive integer.

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According to Euclid’s division lemma, for any positive integer $a$ and a divisor $b=4$, there exist unique integers $q'$ and $r$ such that $a = 4q' + r$, where the remainder $r$ must satisfy the condition $0 \leq r < 4$.

---

The possible integer values for the remainder $r$ are 0, 1, 2, and 3.

Thus, any positive integer $a$ can be expressed in one of the following four forms, for some integer $q' \geq 0$:

• $a = 4q'$
• $a = 4q' + 1$
• $a = 4q' + 2$
• $a = 4q' + 3$

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Now, let's find the square of $a$ in each of these cases:

Case 1: $a = 4q'$

$a^2 = (4q')^2 = 16(q')^2$

We can write this as $a^2 = 4 \times (4(q')^2)$. Let $q = 4(q')^2$. Since $q'$ is an integer, $q$ is also an integer.

So, in this case, $a^2$ is of the form $4q$ for some integer $q$.

---

Case 2: $a = 4q' + 1$

$a^2 = (4q' + 1)^2$

Expand the expression:

$a^2 = (4q')^2 + 2(4q')(1) + 1^2 = 16(q')^2 + 8q' + 1$

Factor out 4 from the first two terms:

$a^2 = 4(4(q')^2 + 2q') + 1$

Let $q = 4(q')^2 + 2q'$. Since $q'$ is an integer, $q$ is also an integer.

So, in this case, $a^2$ is of the form $4q + 1$ for some integer $q$.

---

Case 3: $a = 4q' + 2$

$a^2 = (4q' + 2)^2$

Expand the expression:

$a^2 = (4q')^2 + 2(4q')(2) + 2^2 = 16(q')^2 + 16q' + 4$

Factor out 4 from all terms:

$a^2 = 4(4(q')^2 + 4q' + 1)$

Let $q = 4(q')^2 + 4q' + 1$. Since $q'$ is an integer, $q$ is also an integer.

So, in this case, $a^2$ is of the form $4q$ for some integer $q$.

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Case 4: $a = 4q' + 3$

$a^2 = (4q' + 3)^2$

Expand the expression:

$a^2 = (4q')^2 + 2(4q')(3) + 3^2 = 16(q')^2 + 24q' + 9$

Rewrite the term 9 as $8+1$:

$a^2 = 16(q')^2 + 24q' + 8 + 1$

Factor out 4 from the first three terms:

$a^2 = 4(4(q')^2 + 6q' + 2) + 1$

Let $q = 4(q')^2 + 6q' + 2$. Since $q'$ is an integer, $q$ is also an integer.

So, in this case, $a^2$ is of the form $4q + 1$ for some integer $q$.

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From all the possible cases, we have shown that the square of any positive integer is always of the form $4q$ or $4q + 1$, where $q$ is an integer.

This completes the proof.

To Show: The cube of any positive integer is of the form $4m$, $4m + 1$, or $4m + 3$ for some integer $m$.

---

Proof:

Let $a$ be any positive integer.

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According to Euclid’s division lemma, for any positive integer $a$ and a divisor $b=4$, there exist unique integers $q$ and $r$ such that $a = 4q + r$, where the remainder $r$ must satisfy the condition $0 \leq r < 4$.

---

The possible integer values for the remainder $r$ are 0, 1, 2, and 3.

Thus, any positive integer $a$ can be expressed in one of the following four forms, for some integer $q \geq 0$:

• $a = 4q$
• $a = 4q + 1$
• $a = 4q + 2$
• $a = 4q + 3$

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Now, let's find the cube of $a$ in each of these cases:

Case 1: $a = 4q$

$a^3 = (4q)^3 = 64q^3$

We can write this as $a^3 = 4 \times (16q^3)$. Let $m = 16q^3$. Since $q$ is an integer, $m$ is also an integer.

So, in this case, $a^3$ is of the form $4m$ for some integer $m$.

---

Case 2: $a = 4q + 1$

$a^3 = (4q + 1)^3$

Expand the expression using the formula $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$:

$a^3 = (4q)^3 + 3(4q)^2(1) + 3(4q)(1)^2 + 1^3$

$a^3 = 64q^3 + 3(16q^2) + 12q + 1$

$a^3 = 64q^3 + 48q^2 + 12q + 1$

Factor out 4 from the first three terms:

$a^3 = 4(16q^3 + 12q^2 + 3q) + 1$

Let $m = 16q^3 + 12q^2 + 3q$. Since $q$ is an integer, $m$ is also an integer.

So, in this case, $a^3$ is of the form $4m + 1$ for some integer $m$.

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Case 3: $a = 4q + 2$

$a^3 = (4q + 2)^3$

Expand the expression using the formula $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$:

$a^3 = (4q)^3 + 3(4q)^2(2) + 3(4q)(2)^2 + 2^3$

$a^3 = 64q^3 + 3(16q^2)(2) + 3(4q)(4) + 8$

$a^3 = 64q^3 + 96q^2 + 48q + 8$

Factor out 4 from all terms:

$a^3 = 4(16q^3 + 24q^2 + 12q + 2)$

Let $m = 16q^3 + 24q^2 + 12q + 2$. Since $q$ is an integer, $m$ is also an integer.

So, in this case, $a^3$ is of the form $4m$ for some integer $m$.

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Case 4: $a = 4q + 3$

$a^3 = (4q + 3)^3$

Expand the expression using the formula $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$:

$a^3 = (4q)^3 + 3(4q)^2(3) + 3(4q)(3)^2 + 3^3$

$a^3 = 64q^3 + 3(16q^2)(3) + 3(4q)(9) + 27$

$a^3 = 64q^3 + 144q^2 + 108q + 27$

Rewrite the term 27 as $24 + 3$:

$a^3 = 64q^3 + 144q^2 + 108q + 24 + 3$

Factor out 4 from the first four terms:

$a^3 = 4(16q^3 + 36q^2 + 27q + 6) + 3$

Let $m = 16q^3 + 36q^2 + 27q + 6$. Since $q$ is an integer, $m$ is also an integer.

So, in this case, $a^3$ is of the form $4m + 3$ for some integer $m$.

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From all the possible cases, we have shown that the cube of any positive integer is always of the form $4m$, $4m + 1$, or $4m + 3$, where $m$ is an integer.

This completes the proof.

To Show: The square of any positive integer cannot be of the form $5q + 2$ or $5q + 3$ for any integer $q$.

---

Proof:

Let $a$ be any positive integer.

---

According to Euclid’s division lemma, for any positive integer $a$ and a divisor $b=5$, there exist unique integers $k$ and $r$ such that $a = 5k + r$, where the remainder $r$ must satisfy the condition $0 \leq r < 5$.

---

The possible integer values for the remainder $r$ are 0, 1, 2, 3, and 4.

Thus, any positive integer $a$ can be expressed in one of the following five forms, for some integer $k \geq 0$:

• $a = 5k$
• $a = 5k + 1$
• $a = 5k + 2$
• $a = 5k + 3$
• $a = 5k + 4$

---

Now, let's find the square of $a$ in each of these cases and examine its form when divided by 5:

Case 1: $a = 5k$

$a^2 = (5k)^2 = 25k^2 = 5(5k^2)$

This is of the form $5q$, where $q = 5k^2$. The remainder when $a^2$ is divided by 5 is 0.

---

Case 2: $a = 5k + 1$

$a^2 = (5k + 1)^2 = (5k)^2 + 2(5k)(1) + 1^2 = 25k^2 + 10k + 1 = 5(5k^2 + 2k) + 1$

This is of the form $5q + 1$, where $q = 5k^2 + 2k$. The remainder when $a^2$ is divided by 5 is 1.

---

Case 3: $a = 5k + 2$

$a^2 = (5k + 2)^2 = (5k)^2 + 2(5k)(2) + 2^2 = 25k^2 + 20k + 4 = 5(5k^2 + 4k) + 4$

This is of the form $5q + 4$, where $q = 5k^2 + 4k$. The remainder when $a^2$ is divided by 5 is 4.

---

Case 4: $a = 5k + 3$

$a^2 = (5k + 3)^2 = (5k)^2 + 2(5k)(3) + 3^2 = 25k^2 + 30k + 9$

Rewrite 9 as $5 + 4$:

$a^2 = 25k^2 + 30k + 5 + 4 = 5(5k^2 + 6k + 1) + 4$

This is of the form $5q + 4$, where $q = 5k^2 + 6k + 1$. The remainder when $a^2$ is divided by 5 is 4.

---

Case 5: $a = 5k + 4$

$a^2 = (5k + 4)^2 = (5k)^2 + 2(5k)(4) + 4^2 = 25k^2 + 40k + 16$

Rewrite 16 as $15 + 1$:

$a^2 = 25k^2 + 40k + 15 + 1 = 5(5k^2 + 8k + 3) + 1$

This is of the form $5q + 1$, where $q = 5k^2 + 8k + 3$. The remainder when $a^2$ is divided by 5 is 1.

---

From all the possible cases, we observe that the square of any positive integer, when divided by 5, leaves a remainder of either 0, 1, or 4.

The forms $5q + 2$ and $5q + 3$ correspond to remainders of 2 and 3 respectively when divided by 5.

---

Since the square of any positive integer never leaves a remainder of 2 or 3 when divided by 5, it cannot be of the form $5q + 2$ or $5q + 3$ for any integer $q$.

This completes the proof.

To Show: The square of any positive integer cannot be of the form $6m + 2$ or $6m + 5$ for any integer $m$.

---

Proof:

Let $a$ be any positive integer.

---

According to Euclid’s division lemma, for any positive integer $a$ and a divisor $b=6$, there exist unique integers $k$ and $r$ such that $a = 6k + r$, where the remainder $r$ must satisfy the condition $0 \leq r < 6$.

---

The possible integer values for the remainder $r$ are 0, 1, 2, 3, 4, and 5.

Thus, any positive integer $a$ can be expressed in one of the following six forms, for some integer $k \geq 0$:

• $a = 6k$
• $a = 6k + 1$
• $a = 6k + 2$
• $a = 6k + 3$
• $a = 6k + 4$
• $a = 6k + 5$

---

Now, let's find the square of $a$ in each of these cases and examine its form when divided by 6:

Case 1: $a = 6k$

$a^2 = (6k)^2 = 36k^2 = 6(6k^2)$

This is of the form $6m$, where $m = 6k^2$. The remainder when $a^2$ is divided by 6 is 0.

---

Case 2: $a = 6k + 1$

$a^2 = (6k + 1)^2 = (6k)^2 + 2(6k)(1) + 1^2 = 36k^2 + 12k + 1 = 6(6k^2 + 2k) + 1$

This is of the form $6m + 1$, where $m = 6k^2 + 2k$. The remainder when $a^2$ is divided by 6 is 1.

---

Case 3: $a = 6k + 2$

$a^2 = (6k + 2)^2 = (6k)^2 + 2(6k)(2) + 2^2 = 36k^2 + 24k + 4 = 6(6k^2 + 4k) + 4$

This is of the form $6m + 4$, where $m = 6k^2 + 4k$. The remainder when $a^2$ is divided by 6 is 4.

---

Case 4: $a = 6k + 3$

$a^2 = (6k + 3)^2 = (6k)^2 + 2(6k)(3) + 3^2 = 36k^2 + 36k + 9$

Rewrite 9 as $6 + 3$:

$a^2 = 36k^2 + 36k + 6 + 3 = 6(6k^2 + 6k + 1) + 3$

This is of the form $6m + 3$, where $m = 6k^2 + 6k + 1$. The remainder when $a^2$ is divided by 6 is 3.

---

Case 5: $a = 6k + 4$

$a^2 = (6k + 4)^2 = (6k)^2 + 2(6k)(4) + 4^2 = 36k^2 + 48k + 16$

Rewrite 16 as $12 + 4$:

$a^2 = 36k^2 + 48k + 12 + 4 = 6(6k^2 + 8k + 2) + 4$

This is of the form $6m + 4$, where $m = 6k^2 + 8k + 2$. The remainder when $a^2$ is divided by 6 is 4.

---

Case 6: $a = 6k + 5$

$a^2 = (6k + 5)^2 = (6k)^2 + 2(6k)(5) + 5^2 = 36k^2 + 60k + 25$

Rewrite 25 as $24 + 1$:

$a^2 = 36k^2 + 60k + 24 + 1 = 6(6k^2 + 10k + 4) + 1$

This is of the form $6m + 1$, where $m = 6k^2 + 10k + 4$. The remainder when $a^2$ is divided by 6 is 1.

---

From all the possible cases, we see that the square of any positive integer, when divided by 6, leaves a remainder of either 0, 1, 3, or 4.

The possible forms for the square of a positive integer are $6m$, $6m+1$, $6m+3$, and $6m+4$ for some integer $m$.

---

The forms $6m + 2$ and $6m + 5$ correspond to leaving a remainder of 2 and 5 respectively when divided by 6.

Since the square of any positive integer never leaves a remainder of 2 or 5 when divided by 6, it cannot be of the form $6m + 2$ or $6m + 5$ for any integer $m$.

This completes the proof.

To Show: The square of any odd integer is of the form $4q + 1$ for some integer $q$.

---

Proof:

Let $n$ be an odd integer.

---

By definition, any odd integer can be expressed in the form $2k + 1$, where $k$ is an integer.

So, let $n = 2k + 1$ for some integer $k \in \mathbb{Z}$.

---

Now, consider the square of $n$:

$n^2 = (2k + 1)^2$

---

Expand the expression using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$n^2 = (2k)^2 + 2(2k)(1) + 1^2$

$n^2 = 4k^2 + 4k + 1$

---

We can factor out 4 from the first two terms:

$n^2 = 4(k^2 + k) + 1$

---

Let $q = k^2 + k$. Since $k$ is an integer, $k^2$ is an integer, and the sum of two integers ($k^2$ and $k$) is always an integer. Therefore, $q = k^2 + k$ is an integer.

---

Substituting $q$ back into the expression for $n^2$, we get:

$n^2 = 4q + 1$

---

This shows that the square of any odd integer $n$ is always of the form $4q + 1$, where $q = k^2 + k$ is an integer.

---

This completes the proof.

To Show: If $n$ is an odd integer, then $n^2 – 1$ is divisible by 8.

---

Proof:

Let $n$ be an odd integer.

---

By definition, any odd integer can be expressed in the form $2k + 1$, where $k$ is an integer.

So, let $n = 2k + 1$ for some integer $k \in \mathbb{Z}$.

---

Consider the expression $n^2 - 1$:

$n^2 - 1 = (2k + 1)^2 - 1$

---

Expand the term $(2k + 1)^2$:

$(2k + 1)^2 = (2k)^2 + 2(2k)(1) + 1^2 = 4k^2 + 4k + 1$

---

Substitute this back into the expression for $n^2 - 1$:

$n^2 - 1 = (4k^2 + 4k + 1) - 1$

$n^2 - 1 = 4k^2 + 4k$

---

Factor out $4k$ from the expression:

$n^2 - 1 = 4k(k + 1)$

---

Now, consider the product of consecutive integers $k(k + 1)$. We know that the product of any two consecutive integers is always an even number, because one of the integers must be even.

Thus, $k(k + 1)$ is always divisible by 2.

We can write $k(k + 1) = 2m$ for some integer $m$.

---

Substitute $k(k + 1) = 2m$ into the expression for $n^2 - 1$:

$n^2 - 1 = 4(2m)$

$n^2 - 1 = 8m$

---

Since $n^2 - 1$ can be expressed in the form $8m$, where $m$ is an integer, this means that $n^2 - 1$ is divisible by 8.

---

This holds true for any odd integer $n$.

This completes the proof.

Given: $x$ and $y$ are both odd positive integers.

---

To Prove: $x^2 + y^2$ is even but not divisible by 4.

---

Proof:

Since $x$ and $y$ are odd positive integers, they can be written in the form $2k + 1$ for some non-negative integers $k$.

Let $x = 2k_1 + 1$ and $y = 2k_2 + 1$, where $k_1$ and $k_2$ are integers and $k_1 \geq 0$, $k_2 \geq 0$ (since $x$ and $y$ are positive).

---

Consider the expression $x^2 + y^2$:

$x^2 + y^2 = (2k_1 + 1)^2 + (2k_2 + 1)^2$

---

Expand the squares:

$(2k_1 + 1)^2 = (2k_1)^2 + 2(2k_1)(1) + 1^2 = 4k_1^2 + 4k_1 + 1$

$(2k_2 + 1)^2 = (2k_2)^2 + 2(2k_2)(1) + 1^2 = 4k_2^2 + 4k_2 + 1$

---

Substitute these back into the expression for $x^2 + y^2$:

$x^2 + y^2 = (4k_1^2 + 4k_1 + 1) + (4k_2^2 + 4k_2 + 1)$

$x^2 + y^2 = 4k_1^2 + 4k_1 + 4k_2^2 + 4k_2 + 2$

---

Factor out 4 from the first four terms:

$x^2 + y^2 = 4(k_1^2 + k_1 + k_2^2 + k_2) + 2$

---

Let $m = k_1^2 + k_1 + k_2^2 + k_2$. Since $k_1$ and $k_2$ are integers, $k_1^2$, $k_1$, $k_2^2$, and $k_2$ are integers. The sum of integers is an integer. Therefore, $m$ is an integer.

---

So, the expression for $x^2 + y^2$ becomes:

$x^2 + y^2 = 4m + 2$

---

Now, we analyze the form $4m + 2$:

1. Is $x^2 + y^2$ even?

The expression $4m + 2$ can be written as $2(2m + 1)$. Since $m$ is an integer, $2m + 1$ is also an integer. This shows that $x^2 + y^2$ is a multiple of 2, and thus it is an even number.

---

2. Is $x^2 + y^2$ divisible by 4?

A number is divisible by 4 if it can be expressed in the form $4q$ for some integer $q$. Our expression is $4m + 2$. When $4m + 2$ is divided by 4, the quotient is $m$ and the remainder is 2.

Since the remainder is 2 (and not 0), the number $x^2 + y^2$ is not divisible by 4.

---

We have shown that if $x$ and $y$ are odd positive integers, $x^2 + y^2$ can be written in the form $4m+2$, which is always even but never divisible by 4.

---

This completes the proof.

Solution:

We need to find the HCF of 441, 567, and 693 using Euclid's division algorithm.

First, we find the HCF of any two numbers, say 693 and 567.

Applying Euclid's division algorithm to 693 and 567:

Since $693 > 567$, we divide 693 by 567:

$693 = 567 \times 1 + 126$

Since the remainder $126 \neq 0$, we apply the division algorithm to the new divisor 567 and the remainder 126:

$567 = 126 \times 4 + 63$

Since the remainder $63 \neq 0$, we apply the division algorithm to the new divisor 126 and the remainder 63:

$126 = 63 \times 2 + 0$

The remainder is now 0. The divisor at this stage is 63.

Therefore, the HCF of 693 and 567 is 63.

---

Now, we find the HCF of the third number, 441, and the HCF of the first two numbers, which is 63.

Applying Euclid's division algorithm to 441 and 63:

Since $441 > 63$, we divide 441 by 63:

$441 = 63 \times 7 + 0$

The remainder is now 0. The divisor at this stage is 63.

Therefore, the HCF of 441 and 63 is 63.

---

The HCF of 441, 567, and 693 is the HCF of (HCF(693, 567), 441), which is HCF(63, 441).

Thus, the HCF of 441, 567, and 693 is 63.

Solution:

Let the largest number be $N$.

According to the problem, when 1251 is divided by $N$, the remainder is 1. This implies that $1251 - 1 = 1250$ is exactly divisible by $N$.

When 9377 is divided by $N$, the remainder is 2. This implies that $9377 - 2 = 9375$ is exactly divisible by $N$.

When 15628 is divided by $N$, the remainder is 3. This implies that $15628 - 3 = 15625$ is exactly divisible by $N$.

Thus, the largest number $N$ that divides 1251, 9377, and 15628 leaving remainders 1, 2, and 3 respectively, is the HCF of 1250, 9375, and 15625.

---

First, we find the HCF of 9375 and 1250 using Euclid's division algorithm.

Since $9375 > 1250$, we apply the division algorithm:

$9375 = 1250 \times 7 + 625$

Since the remainder $625 \neq 0$, we apply the division algorithm to the new divisor 1250 and the remainder 625:

$1250 = 625 \times 2 + 0$

The remainder is now 0. The divisor at this stage is 625.

Therefore, the HCF of 9375 and 1250 is 625.

---

Now, we find the HCF of the third number 15625 and the HCF of the first two numbers, which is 625.

Applying Euclid's division algorithm to 15625 and 625:

Since $15625 > 625$, we apply the division algorithm:

$15625 = 625 \times 25 + 0$

The remainder is now 0. The divisor at this stage is 625.

Therefore, the HCF of 15625 and 625 is 625.

---

The HCF of 1250, 9375, and 15625 is the HCF of (HCF(9375, 1250), 15625), which is HCF(625, 15625).

Thus, the HCF of 1250, 9375, and 15625 is 625.

The largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively is 625.

Proof:

Let us assume, to the contrary, that $\sqrt{3} + \sqrt{5}$ is a rational number.

If $\sqrt{3} + \sqrt{5}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p, q$ are coprime (having no common factors other than 1).

So, we have:

$\sqrt{3} + \sqrt{5} = \frac{p}{q}$

Rearranging the terms, we can isolate one of the square roots. Let's isolate $\sqrt{3}$:

$\sqrt{3} = \frac{p}{q} - \sqrt{5}$

Squaring both sides of the equation:

$(\sqrt{3})^2 = \left(\frac{p}{q} - \sqrt{5}\right)^2$

$3 = \left(\frac{p}{q}\right)^2 - 2 \left(\frac{p}{q}\right) \sqrt{5} + (\sqrt{5})^2$

$3 = \frac{p^2}{q^2} - \frac{2p}{q}\sqrt{5} + 5$

Now, rearrange the equation to isolate the term with $\sqrt{5}$:

$\frac{2p}{q}\sqrt{5} = \frac{p^2}{q^2} + 5 - 3$

$\frac{2p}{q}\sqrt{5} = \frac{p^2}{q^2} + 2$

Combine the terms on the right side:

$\frac{2p}{q}\sqrt{5} = \frac{p^2 + 2q^2}{q^2}$

Now, isolate $\sqrt{5}$:

$\sqrt{5} = \frac{p^2 + 2q^2}{q^2} \times \frac{q}{2p}$

$\sqrt{5} = \frac{p^2 + 2q^2}{2pq}$

---

Since $p$ and $q$ are integers, the expression $\frac{p^2 + 2q^2}{2pq}$ is a rational number (as long as $2pq \neq 0$, which is true because $q \neq 0$ and if $p=0$, then $\sqrt{3}+\sqrt{5}=0$, which is false).

So, the equation $\sqrt{5} = \frac{p^2 + 2q^2}{2pq}$ implies that $\sqrt{5}$ is a rational number.

However, we know that $\sqrt{5}$ is an irrational number.

---

This leads to a contradiction: a number cannot be both rational and irrational.

This contradiction arose because of our initial assumption that $\sqrt{3} + \sqrt{5}$ is rational.

Therefore, our assumption was false, and $\sqrt{3} + \sqrt{5}$ must be an irrational number.

Hence, proved.

To Prove:

That $12^n$ cannot end with the digit 0 or 5 for any natural number $n$.

---

Proof:

We know that any positive integer that ends with the digit 0 or 5 must be divisible by 5.

For a number to be divisible by 5, its prime factorization must contain the prime number 5.

Let's find the prime factorization of the base number, 12.

The prime factorization of 12 is:

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

So, $12 = 2 \times 2 \times 3 = 2^2 \times 3$.

Now, consider the number $12^n$ for any natural number $n$.

The prime factorization of $12^n$ is obtained by raising the prime factors of 12 to the power of $n$:

$12^n = (2^2 \times 3)^n = (2^2)^n \times 3^n = 2^{2n} \times 3^n$

---

The prime factors in the factorization of $12^n$ are only 2 and 3. There is no prime factor of 5 in the factorization of $12^n$.

According to the Fundamental Theorem of Arithmetic, the prime factorization of every composite number is unique.

This uniqueness means that the only prime numbers that can divide $12^n$ are 2 and 3.

Since 5 is not a prime factor of $12^n$, $12^n$ is not divisible by 5.

Therefore, $12^n$ cannot end with the digit 0 (which requires divisibility by both 2 and 5) or the digit 5 (which requires divisibility by 5) for any natural number $n$.

Hence, proved.

To Find:

The minimum distance each person should walk so that each can cover the same distance in complete steps.

---

Solution:

The distance covered by each person must be a multiple of their respective step lengths (40 cm, 42 cm, and 45 cm).

To find the minimum distance that can be covered by all three in complete steps, we need to find the Least Common Multiple (LCM) of 40, 42, and 45.

We will find the LCM using the prime factorization method or the common division method.

Let's use the common division method:

$\begin{array}{c|cc} 2 & 40 \;, & 42 \;, & 45 \\ \hline 2 & 20 \; , & 21 \; , & 45 \\ \hline 2 & 10 \; , & 21 \; , & 45 \\ \hline 3 & 5 \; , & 21 \; , & 45 \\ \hline 3 & 5 \; , & 7 \; , & 15 \\ \hline 5 & 5 \; , & 7 \; , & 5 \\ \hline 7 & 1 \; , & 7 \; , & 1 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

The LCM is the product of the prime factors:

LCM$(40, 42, 45) = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7$

LCM$(40, 42, 45) = 2^3 \times 3^2 \times 5 \times 7$

LCM$(40, 42, 45) = 8 \times 9 \times 5 \times 7$

LCM$(40, 42, 45) = 72 \times 35$

Calculating the product:

$72 \times 35 = 2520$

---

The LCM of 40, 42, and 45 is 2520.

This means that the minimum distance each person should walk so that each can cover the same distance in complete steps is 2520 cm.

Note: 2520 cm is equal to 25.2 meters.

Solution:

The given rational number is $\frac{257}{5000}$.

The denominator of this rational number is 5000.

---

First, we find the prime factorization of the denominator, 5000.

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

So, the prime factorization of 5000 is $2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = 2^3 \times 5^4$.

We can write the denominator in the form $2^m \times 5^n$ as:

$5000 = 2^3 \times 5^4$

Here, $m = 3$ and $n = 4$, which are non-negative integers.

---

Now, we write the decimal expansion of $\frac{257}{5000}$ without actual division.

We have the fraction $\frac{257}{2^3 \times 5^4}$.

To express this as a decimal easily, we need the denominator to be a power of 10. A power of 10 is of the form $10^k = (2 \times 5)^k = 2^k \times 5^k$.

In our denominator $2^3 \times 5^4$, the highest power is 4 (from $5^4$). To make the powers of 2 and 5 equal to 4, we need to multiply by $2^{4-3} = 2^1 = 2$.

We multiply both the numerator and the denominator by 2:

$\frac{257}{2^3 \times 5^4} = \frac{257 \times 2}{2^3 \times 5^4 \times 2^1}$

$\frac{257 \times 2}{2^{3+1} \times 5^4} = \frac{514}{2^4 \times 5^4}$

Now, the denominator is $2^4 \times 5^4 = (2 \times 5)^4 = 10^4 = 10000$.

So, the fraction becomes $\frac{514}{10000}$.

To convert a fraction with a denominator of 10000 to a decimal, we place the decimal point 4 places from the right (since there are 4 zeros in 10000).

$\frac{514}{10000} = 0.0514$

---

The denominator of $\frac{257}{5000}$ in the form $2^m \times 5^n$ is $2^3 \times 5^4$, where $m=3$ and $n=4$.

The decimal expansion of $\frac{257}{5000}$ is 0.0514.

To Prove:

That $\sqrt{p} + \sqrt{q}$ is irrational, where $p$ and $q$ are prime numbers.

---

Proof:

Let us assume, to the contrary, that $\sqrt{p} + \sqrt{q}$ is a rational number.

If $\sqrt{p} + \sqrt{q}$ is rational, then it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a, b$ are coprime (having no common factors other than 1).

So, we have:

$\sqrt{p} + \sqrt{q} = \frac{a}{b}$

Rearranging the terms, we can isolate one of the square roots. Let's isolate $\sqrt{p}$:

$\sqrt{p} = \frac{a}{b} - \sqrt{q}$

Squaring both sides of the equation:

$(\sqrt{p})^2 = \left(\frac{a}{b} - \sqrt{q}\right)^2$

$p = \left(\frac{a}{b}\right)^2 - 2 \left(\frac{a}{b}\right) \sqrt{q} + (\sqrt{q})^2$

$p = \frac{a^2}{b^2} - \frac{2a}{b}\sqrt{q} + q$

Now, rearrange the equation to isolate the term with $\sqrt{q}$:

$\frac{2a}{b}\sqrt{q} = \frac{a^2}{b^2} + q - p$

$\frac{2a}{b}\sqrt{q} = \frac{a^2 + (q-p)b^2}{b^2}$

Now, isolate $\sqrt{q}$. Note that $a \neq 0$, because if $a=0$, then $\sqrt{p} + \sqrt{q} = 0$, which is impossible since $p$ and $q$ are positive primes.

$\sqrt{q} = \frac{a^2 + (q-p)b^2}{b^2} \times \frac{b}{2a}$

$\sqrt{q} = \frac{a^2 + (q-p)b^2}{2ab}$

---

Since $a$, $b$, $p$, and $q$ are integers, the expression $\frac{a^2 + (q-p)b^2}{2ab}$ is a rational number (as $2ab \neq 0$).

So, the equation $\sqrt{q} = \frac{a^2 + (q-p)b^2}{2ab}$ implies that $\sqrt{q}$ is a rational number.

However, we know that $\sqrt{q}$ is an irrational number because $q$ is a prime number.

---

This leads to a contradiction: a number cannot be both rational and irrational.

This contradiction arose because of our initial assumption that $\sqrt{p} + \sqrt{q}$ is rational.

Therefore, our assumption was false, and $\sqrt{p} + \sqrt{q}$ must be an irrational number.

Hence, proved.

To Prove:

The square of an odd positive integer can be of the form $6q + 1$ or $6q + 3$ for some integer $q$.

---

Proof:

Let $a$ be an odd positive integer.

According to Euclid's division algorithm, any integer $a$ can be expressed in the form $6k + r$, where $k$ is an integer ($k \geq 0$ for a positive integer) and $r$ is the remainder, where $0 \leq r < 6$.

So, possible forms for $a$ are $6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5$.

Since $a$ is an odd positive integer, it cannot be of the form $6k$ (divisible by 2), $6k+2$ (divisible by 2), or $6k+4$ (divisible by 2).

Thus, an odd positive integer $a$ must be of the form $6k+1$, $6k+3$, or $6k+5$ for some integer $k \geq 0$.

---

Now, let's consider the square of each of these forms:

Case 1: $a = 6k + 1$

$a^2 = (6k+1)^2$

$a^2 = (6k)^2 + 2(6k)(1) + 1^2$

$a^2 = 36k^2 + 12k + 1$

$a^2 = 6(6k^2 + 2k) + 1$

Let $q = 6k^2 + 2k$. Since $k$ is an integer, $q$ is also an integer.

So, $a^2 = 6q + 1$.

---

Case 2: $a = 6k + 3$

$a^2 = (6k+3)^2$

$a^2 = (6k)^2 + 2(6k)(3) + 3^2$

$a^2 = 36k^2 + 36k + 9$

$a^2 = 36k^2 + 36k + 6 + 3$

$a^2 = 6(6k^2 + 6k + 1) + 3$

Let $q = 6k^2 + 6k + 1$. Since $k$ is an integer, $q$ is also an integer.

So, $a^2 = 6q + 3$.

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Case 3: $a = 6k + 5$

$a^2 = (6k+5)^2$

$a^2 = (6k)^2 + 2(6k)(5) + 5^2$

$a^2 = 36k^2 + 60k + 25$

$a^2 = 36k^2 + 60k + 24 + 1$

$a^2 = 6(6k^2 + 10k + 4) + 1$

Let $q = 6k^2 + 10k + 4$. Since $k$ is an integer, $q$ is also an integer.

So, $a^2 = 6q + 1$.

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From all possible cases for an odd positive integer, we see that the square of an odd positive integer is always of the form $6q + 1$ or $6q + 3$ for some integer $q$.

Hence, proved.

To Prove:

The cube of a positive integer of the form $6q + r$, where $q$ is an integer and $r \in \{0, 1, 2, 3, 4, 5\}$, is also of the form $6m + r$ for some integer $m$.

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Proof:

Let $n$ be a positive integer of the form $6q + r$, where $q$ is an integer and $r \in \{0, 1, 2, 3, 4, 5\}$.

We consider the cube of $n$:

$n^3 = (6q + r)^3$

Using the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, we expand $(6q+r)^3$ with $a=6q$ and $b=r$:

$n^3 = (6q)^3 + 3(6q)^2(r) + 3(6q)(r^2) + r^3$

$n^3 = 216q^3 + 3(36q^2)r + 18qr^2 + r^3$

$n^3 = 216q^3 + 108q^2r + 18qr^2 + r^3$

We can factor out 6 from the first three terms:

$n^3 = 6(36q^3 + 18q^2r + 3qr^2) + r^3$

Let $K = 36q^3 + 18q^2r + 3qr^2$. Since $q$ and $r$ are integers, $K$ is also an integer.

So, $n^3 = 6K + r^3$.

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Now, we need to show that $r^3$ can be written in the form $6J + r$ for some integer $J$, for each value of $r \in \{0, 1, 2, 3, 4, 5\}$. This is equivalent to showing that $r^3 - r$ is divisible by 6 for these values of $r$.

Consider the expression $r^3 - r$:

$r^3 - r = r(r^2 - 1) = r(r-1)(r+1)$

The expression $r(r-1)(r+1)$ is the product of three consecutive integers: $(r-1)$, $r$, and $(r+1)$.

The product of any two consecutive integers is divisible by 2.

The product of any three consecutive integers is divisible by 3.

Since the product is divisible by both 2 and 3, and 2 and 3 are coprime, the product must be divisible by $2 \times 3 = 6$.

Thus, $r^3 - r$ is divisible by 6 for any integer $r$. This holds specifically for $r \in \{0, 1, 2, 3, 4, 5\}$.

So, we can write $r^3 - r = 6J_r$ for some integer $J_r$ (which depends on $r$).

This means $r^3 = 6J_r + r$.

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Substitute this back into the expression for $n^3$:

$n^3 = 6K + r^3$

$n^3 = 6K + (6J_r + r)$

$n^3 = 6K + 6J_r + r$

$n^3 = 6(K + J_r) + r$

Let $m = K + J_r$. Since $K$ and $J_r$ are integers, $m$ is also an integer.

Therefore, the cube of a positive integer of the form $6q + r$ is of the form $6m + r$, where $m$ is some integer.

This applies for all values of $r \in \{0, 1, 2, 3, 4, 5\}$.

Hence, proved.

To Prove:

One and only one out of $n$, $n+2$, and $n+4$ is divisible by 3, where $n$ is any positive integer.

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Proof:

Let $n$ be any positive integer.

According to Euclid's division algorithm, when $n$ is divided by 3, the remainder can be 0, 1, or 2.

Thus, $n$ can be of the form $3k$, $3k+1$, or $3k+2$ for some integer $k \geq 0$ (since $n$ is positive).

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We examine the divisibility of $n$, $n+2$, and $n+4$ by 3 for each of these cases:

Case 1: $n = 3k$ for some integer $k \geq 1$ (since $n$ is positive).

If $n=3k$, then:

$n = 3k$. This is clearly divisible by 3.

$n+2 = (3k) + 2 = 3k + 2$. When divided by 3, the remainder is 2. Not divisible by 3.

$n+4 = (3k) + 4 = 3k + 3 + 1 = 3(k+1) + 1$. When divided by 3, the remainder is 1. Not divisible by 3.

In this case, only $n$ is divisible by 3.

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Case 2: $n = 3k + 1$ for some integer $k \geq 0$.

If $n = 3k+1$, then:

$n = 3k + 1$. When divided by 3, the remainder is 1. Not divisible by 3.

$n+2 = (3k + 1) + 2 = 3k + 3 = 3(k+1)$. This is clearly divisible by 3.

$n+4 = (3k + 1) + 4 = 3k + 5 = 3k + 3 + 2 = 3(k+1) + 2$. When divided by 3, the remainder is 2. Not divisible by 3.

In this case, only $n+2$ is divisible by 3.

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Case 3: $n = 3k + 2$ for some integer $k \geq 0$.

If $n = 3k+2$, then:

$n = 3k + 2$. When divided by 3, the remainder is 2. Not divisible by 3.

$n+2 = (3k + 2) + 2 = 3k + 4 = 3k + 3 + 1 = 3(k+1) + 1$. When divided by 3, the remainder is 1. Not divisible by 3.

$n+4 = (3k + 2) + 4 = 3k + 6 = 3(k+2)$. This is clearly divisible by 3.

In this case, only $n+4$ is divisible by 3.

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From the three possible cases for $n$, we see that in each case, exactly one of the numbers $n$, $n+2$, and $n+4$ is divisible by 3.

Therefore, one and only one out of $n$, $n+2$, and $n+4$ is divisible by 3 for any positive integer $n$.

Hence, proved.

To Prove:

One of any three consecutive positive integers must be divisible by 3.

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Proof:

Let the three consecutive positive integers be $n$, $n+1$, and $n+2$, where $n$ is a positive integer.

According to Euclid's division algorithm, when any positive integer $n$ is divided by 3, the remainder can be 0, 1, or 2.

Thus, $n$ can be expressed in one of the following forms for some integer $k \geq 0$:

• $n = 3k$
• $n = 3k + 1$
• $n = 3k + 2$

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We examine the divisibility by 3 for each case:

Case 1: $n = 3k$

If $n = 3k$, then $n$ is clearly divisible by 3.

$n+1 = 3k + 1$. When divided by 3, the remainder is 1.

$n+2 = 3k + 2$. When divided by 3, the remainder is 2.

In this case, $n$ is divisible by 3.

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Case 2: $n = 3k + 1$

If $n = 3k + 1$, then when divided by 3, the remainder is 1.

$n+1 = (3k + 1) + 1 = 3k + 2$. When divided by 3, the remainder is 2.

$n+2 = (3k + 1) + 2 = 3k + 3 = 3(k+1)$. This is clearly divisible by 3.

In this case, $n+2$ is divisible by 3.

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Case 3: $n = 3k + 2$

If $n = 3k + 2$, then when divided by 3, the remainder is 2.

$n+1 = (3k + 2) + 1 = 3k + 3 = 3(k+1)$. This is clearly divisible by 3.

$n+2 = (3k + 2) + 2 = 3k + 4 = 3k + 3 + 1 = 3(k+1) + 1$. When divided by 3, the remainder is 1.

In this case, $n+1$ is divisible by 3.

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In each of the possible cases for the form of $n$, exactly one of the three consecutive integers ($n$, $n+1$, or $n+2$) is divisible by 3.

Therefore, one of any three consecutive positive integers must be divisible by 3.

Hence, proved.

To Prove:

For any positive integer $n$, $n^3 - n$ is divisible by 6.

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Proof:

We need to show that $n^3 - n$ is divisible by 6 for any positive integer $n$.

First, we factorize the expression $n^3 - n$:

$n^3 - n = n(n^2 - 1)$

Using the difference of squares formula, $n^2 - 1 = (n-1)(n+1)$:

$n^3 - n = n(n-1)(n+1)$

Rearranging the terms, we get:

$n^3 - n = (n-1)n(n+1)$

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The expression $(n-1)n(n+1)$ is the product of three consecutive integers: $n-1$, $n$, and $n+1$.

We know the following properties about the product of consecutive integers:

1. The product of any two consecutive integers is always divisible by 2.

In the product $(n-1)n(n+1)$, the terms $(n-1)$ and $n$ are consecutive integers, so their product $(n-1)n$ is divisible by 2. Therefore, the entire product $(n-1)n(n+1)$ is divisible by 2.

2. The product of any three consecutive integers is always divisible by 3.

Out of any three consecutive integers, one of them must be a multiple of 3. Therefore, the product $(n-1)n(n+1)$ is divisible by 3.

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Since $n^3 - n = (n-1)n(n+1)$ is divisible by both 2 and 3, and because 2 and 3 are coprime numbers (their greatest common divisor is 1), their product $2 \times 3 = 6$ must also divide $n^3 - n$.

Therefore, $n^3 - n$ is divisible by 6 for any positive integer $n$.

Hence, proved.

To Prove:

One and only one out of $n$, $n+4$, $n+8$, $n+12$, and $n+16$ is divisible by 5, where $n$ is any positive integer.

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Proof:

Let $n$ be any positive integer.

According to Euclid's division algorithm, when $n$ is divided by 5, the remainder can be 0, 1, 2, 3, or 4.

Thus, $n$ can be expressed in one of the following forms for some integer $k \geq 0$:

• $n = 5k$
• $n = 5k + 1$
• $n = 5k + 2$
• $n = 5k + 3$
• $n = 5k + 4$

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We examine the divisibility by 5 for each case:

Case 1: $n = 5k$ for some integer $k \geq 1$ (since $n$ is positive).

If $n = 5k$, then:

$n = 5k$. This is clearly divisible by 5.

$n+4 = 5k + 4$. When divided by 5, the remainder is 4.

$n+8 = 5k + 8 = 5k + 5 + 3 = 5(k+1) + 3$. When divided by 5, the remainder is 3.

$n+12 = 5k + 12 = 5k + 10 + 2 = 5(k+2) + 2$. When divided by 5, the remainder is 2.

$n+16 = 5k + 16 = 5k + 15 + 1 = 5(k+3) + 1$. When divided by 5, the remainder is 1.

In this case, only $n$ is divisible by 5.

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Case 2: $n = 5k + 1$ for some integer $k \geq 0$.

If $n = 5k+1$, then:

$n = 5k + 1$. When divided by 5, the remainder is 1.

$n+4 = (5k + 1) + 4 = 5k + 5 = 5(k+1)$. This is clearly divisible by 5.

$n+8 = (5k + 1) + 8 = 5k + 9 = 5k + 5 + 4 = 5(k+1) + 4$. When divided by 5, the remainder is 4.

$n+12 = (5k + 1) + 12 = 5k + 13 = 5k + 10 + 3 = 5(k+2) + 3$. When divided by 5, the remainder is 3.

$n+16 = (5k + 1) + 16 = 5k + 17 = 5k + 15 + 2 = 5(k+3) + 2$. When divided by 5, the remainder is 2.

In this case, only $n+4$ is divisible by 5.

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Case 3: $n = 5k + 2$ for some integer $k \geq 0$.

If $n = 5k+2$, then:

$n = 5k + 2$. When divided by 5, the remainder is 2.

$n+4 = (5k + 2) + 4 = 5k + 6 = 5k + 5 + 1 = 5(k+1) + 1$. When divided by 5, the remainder is 1.

$n+8 = (5k + 2) + 8 = 5k + 10 = 5(k+2)$. This is clearly divisible by 5.

$n+12 = (5k + 2) + 12 = 5k + 14 = 5k + 10 + 4 = 5(k+2) + 4$. When divided by 5, the remainder is 4.

$n+16 = (5k + 2) + 16 = 5k + 18 = 5k + 15 + 3 = 5(k+3) + 3$. When divided by 5, the remainder is 3.

In this case, only $n+8$ is divisible by 5.

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Case 4: $n = 5k + 3$ for some integer $k \geq 0$.

If $n = 5k+3$, then:

$n = 5k + 3$. When divided by 5, the remainder is 3.

$n+4 = (5k + 3) + 4 = 5k + 7 = 5k + 5 + 2 = 5(k+1) + 2$. When divided by 5, the remainder is 2.

$n+8 = (5k + 3) + 8 = 5k + 11 = 5k + 10 + 1 = 5(k+2) + 1$. When divided by 5, the remainder is 1.

$n+12 = (5k + 3) + 12 = 5k + 15 = 5(k+3)$. This is clearly divisible by 5.

$n+16 = (5k + 3) + 16 = 5k + 19 = 5k + 15 + 4 = 5(k+3) + 4$. When divided by 5, the remainder is 4.

In this case, only $n+12$ is divisible by 5.

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Case 5: $n = 5k + 4$ for some integer $k \geq 0$.

If $n = 5k+4$, then:

$n = 5k + 4$. When divided by 5, the remainder is 4.

$n+4 = (5k + 4) + 4 = 5k + 8 = 5k + 5 + 3 = 5(k+1) + 3$. When divided by 5, the remainder is 3.

$n+8 = (5k + 4) + 8 = 5k + 12 = 5k + 10 + 2 = 5(k+2) + 2$. When divided by 5, the remainder is 2.

$n+12 = (5k + 4) + 12 = 5k + 16 = 5k + 15 + 1 = 5(k+3) + 1$. When divided by 5, the remainder is 1.

$n+16 = (5k + 4) + 16 = 5k + 20 = 5(k+4)$. This is clearly divisible by 5.

In this case, only $n+16$ is divisible by 5.

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From all five possible cases for the form of $n$, we see that in each case, exactly one of the numbers $n$, $n+4$, $n+8$, $n+12$, and $n+16$ is divisible by 5.

Therefore, one and only one out of $n$, $n+4$, $n+8$, $n+12$, and $n+16$ is divisible by 5, where $n$ is any positive integer.

Hence, proved.