Chapter 1 Real Numbers (Class 10 - Maths NCERT Exemplar Solutions)

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.

Given:

The expression is $n^2 - 1$.

The condition is that $n^2 - 1$ must be divisible by $8$.

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To Find:

The nature of the number $n$ (whether it is an integer, a natural number, an odd integer, or an even integer).

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Solution:

Let us test the expression $n^2 - 1$ for different types of integers:

Case 1: If $n$ is an even integer.

Let $n = 2k$, where $k$ is an integer.

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

For any integer $k$, $4k^2$ is always even, so $4k^2 - 1$ will always be odd. Since an odd number cannot be divisible by $8$, $n$ cannot be an even integer.

Case 2: If $n$ is an odd integer.

Let $n = 2k + 1$, where $k$ is an integer.

Substituting this value in the expression:

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

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

We know that $k$ and $(k + 1)$ are two consecutive integers. The product of any two consecutive integers is always even.

Substituting this into equation (ii):

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

Since $n^2 - 1$ is a multiple of $8$, it is divisible by $8$. This holds true whenever $n$ is an odd integer.

Thus, the correct option is (C).

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Alternate Solution (Substitution Method):

We can check the divisibility by substituting different values of $n$ into the expression $n^2 - 1$ as shown in the table below:

Nature of $n$	Value of $n$	$n^2 - 1$	Divisible by 8?
Odd	1	$1^2 - 1 = 0$	Yes ($0 \div 8 = 0$)
Even	2	$2^2 - 1 = 3$	No
Odd	3	$3^2 - 1 = 8$	Yes
Even	4	$4^2 - 1 = 15$	No
Odd	5	$5^2 - 1 = 24$	Yes

From the table, it is clear that the result is divisible by $8$ only when $n$ is an odd integer ($1, 3, 5, \dots$).

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

Correct Option: (C)

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.

Given:

Where $x$ and $y$ are prime numbers.

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To Find:

The Highest Common Factor, $\text{HCF} (a, b)$.

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Solution:

We know that the Highest Common Factor (HCF) of two or more numbers is the product of the lowest powers of all common prime factors involved in the numbers.

Let us list the prime factors and their powers for both integers $a$ and $b$:

For $a = x^3 y^2$:

The power of prime factor $x$ is $3$.

The power of prime factor $y$ is $2$.

For $b = x y^3$:

The power of prime factor $x$ is $1$.

The power of prime factor $y$ is $3$.

To find the HCF, we select the common prime factors with the smallest exponents:

Therefore, the HCF is obtained by multiplying these lowest powers:

$\text{HCF} (a, b) = xy^2$

Hence, the correct option is (B) $xy^2$.

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Comparison Table:

Prime Factor	Power in $a$ ($x^3 y^2$)	Power in $b$ ($x y^3$)	Lowest Power (HCF)
$x$	3	1	$x^1$
$y$	2	3	$y^2$

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Alternate Solution:

We can write the prime factorisation of $a$ and $b$ in expanded form:

$a = x \times x \times x \times y \times y$

$b = x \times y \times y \times y$

The common factors in both $a$ and $b$ are:

One '$x$' and two '$y$'s ($y \times y$).

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

$\text{HCF} (a, b) = xy^2$

This confirms that the HCF of $a$ and $b$ is $xy^2$.

Correct Option: (B)

Given:

where $a$ and $b$ are prime numbers.

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To Find:

The Least Common Multiple, $\text{LCM} (p, q)$.

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Solution:

We know that the Least Common Multiple (LCM) of two or more numbers is the product of the highest powers of all prime factors involved in the numbers.

Let us identify the prime factors and their highest exponents in the expressions of $p$ and $q$:

For $p = ab^2$:

The power of prime factor $a$ is $1$.

The power of prime factor $b$ is $2$.

For $q = a^3 b$:

The power of prime factor $a$ is $3$.

The power of prime factor $b$ is $1$.

To find the LCM, we select the highest power for each prime factor present:

Therefore, the LCM is the product of these highest powers:

$\text{LCM} (p, q) = a^3 b^2$

Hence, the correct option is (C) $a^3b^2$.

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Comparison Table:

Prime Factor	Power in $p$ ($ab^2$)	Power in $q$ ($a^3b$)	Highest Power (LCM)
$a$	1	3	$a^3$
$b$	2	1	$b^2$

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Alternate Solution:

We can use the relationship between HCF and LCM for two positive integers:

First, find the HCF of $p = ab^2$ and $q = a^3b$ by taking the lowest powers of common factors:

$\text{HCF} (p, q) = a^1 \times b^1 = ab$

Now, substitute the values in the formula:

$(ab) \times \text{LCM} (p, q) = (ab^2) \times (a^3b)$

$(ab) \times \text{LCM} (p, q) = a^4 b^3$

$\text{LCM} (p, q) = a^{(4-1)} b^{(3-1)}$

$\text{LCM} (p, q) = a^3 b^2$

This confirms that the LCM of $p$ and $q$ is $a^3b^2$.

Correct Option: (C)

Given:

A non-zero rational number and an irrational number.

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To Find:

The nature of the product of a non-zero rational number and an irrational number.

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Solution:

According to the fundamental properties of real numbers, the product of any non-zero rational number and an irrational number is always irrational.

Let us understand this with the help of examples in the table below:

Non-zero Rational ($r$)	Irrational Number ($i$)	Product ($r \times i$)	Nature of Result
$2$	$\sqrt{3}$	$2\sqrt{3}$	Irrational
$\frac{1}{2}$	$\pi$	$\frac{\pi}{2}$	Irrational
$-5$	$\sqrt{7}$	$-5\sqrt{7}$	Irrational

If the rational number were zero, the product would be $0 \times \text{irrational} = 0$, which is rational. However, the question specifically mentions a non-zero rational number.

Therefore, the product is always irrational.

Hence, the correct option is (A).

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Alternate Solution (Proof by Contradiction):

Let $r$ be a non-zero rational number and $x$ be an irrational number. Let us assume, for the sake of contradiction, that their product $P$ is a rational number.

Since $P$ is assumed to be rational and $r$ is a given non-zero rational, we can write:

We know that the division of one rational number by another non-zero rational number always results in a rational number.

Substituting this back into equation (ii), we get:

$x = \text{Rational Number}$

But this contradicts the given fact that $x$ is an irrational number. Our assumption that the product is rational must be false.

Thus, the product of a non-zero rational and an irrational number is always irrational.

Correct Option: (A)

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.

Given:

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

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To Find:

The number of decimal places after which the decimal expansion of $\frac{14587}{1250}$ will terminate.

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Solution:

A rational number $\frac{p}{q}$ (where $p$ and $q$ are co-prime) has a terminating decimal expansion if the prime factorisation of the denominator $q$ is of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers. The decimal expansion terminates after $k$ places, where $k = \max(m, n)$.

First, let us perform the prime factorisation of the denominator $1250$:

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

From the prime factorisation, we can write the denominator as:

Comparing $2^1 \times 5^4$ with the standard form $2^m \times 5^n$, we get:

$m = 1$ and $n = 4$.

The number of decimal places after which the expansion terminates is given by the highest power among $m$ and $n$.

Therefore, the decimal expansion of $\frac{14587}{1250}$ will terminate after four decimal places.

Hence, the correct option is (D).

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Alternate Solution:

We can find the decimal expansion by making the denominator a power of $10$. To do this, we multiply both the numerator and the denominator by a suitable power of $2$ to match the power of $5$.

Multiply the numerator and denominator by $2^3$:

$\frac{14587 \times 2^3}{2^1 \times 5^4 \times 2^3} = \frac{14587 \times 8}{2^4 \times 5^4}$

$\frac{116696}{(2 \times 5)^4} = \frac{116696}{10^4}$

In the decimal form $11.6696$, there are four digits after the decimal point.

Thus, the decimal expansion terminates after four decimal places.

Correct Option: (D)

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 $\{0, 1\}$.

Given:

The expression is $6^n$, where $n$ is a natural number (i.e., $n = 1, 2, 3, \dots$).

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To Find:

Whether the number $6^n$ can end with the digit $5$ for any natural number $n$.

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Solution:

If any number ends with the digit $5$, it must be divisible by $5$. This means that the prime factorisation of that number must contain the prime factor $5$.

Let us examine the prime factorisation of the base $6$:

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

So, the prime factorisation of $6$ is:

Now, for $6^n$, we have:

The prime factors of $6^n$ are only $2$ and $3$. According to the Fundamental Theorem of Arithmetic, the prime factorisation of every composite number is unique.

Since the prime factor $5$ does not appear in the prime factorisation of $6^n$, the number $6^n$ is not divisible by 5.

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

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Alternate Solution:

We can observe the pattern of the last digit (units digit) of $6^n$ by substituting various values of $n$ as shown in the table below:

Value of $n$	Expression $6^n$	Result	Ends with digit
1	$6^1$	6	6
2	$6^2$	36	6
3	$6^3$	216	6
4	$6^4$	1296	6

From the table, it is evident that for any natural number $n$, $6^n$ always ends with the digit $6$.

In fact, any number ending in $6$, when multiplied by $6$, will again result in a number ending in $6$.

Since the units digit is always $6$, it can never be $5$.

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Conclusion:

No, the number $6^n$ cannot end with the digit $5$ because its prime factorisation does not contain $5$ and its units digit always remains $6$.

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.

Given:

A positive integer form $3m + 2$, where $m$ is a natural number.

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To Find:

Whether the square of any positive integer can be of the form $3m + 2$.

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Solution:

According to Euclid’s Division Lemma, any positive integer $n$ can be expressed in the form $3q + r$, where $q$ is some integer and $r$ is the remainder.

This implies that any positive integer can be of the form $3q$, $3q + 1$ or $3q + 2$. Let us square these integers one by one:

Case I: When $n = 3q$

Squaring both sides:

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

Case II: When $n = 3q + 1$

Squaring both sides:

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

Taking $3$ as a common factor from the first two terms:

Case III: When $n = 3q + 2$

Squaring both sides:

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

$n^2 = 9q^2 + 12q + 4$

We can rewrite $4$ as $3 + 1$ to take $3$ as a common factor:

$n^2 = 9q^2 + 12q + 3 + 1$

From equations (i), (ii), and (iii), it is clear that the square of any positive integer is either of the form $3m$ or $3m + 1$, but never of the form $3m + 2$.

Conclusion: No, the square of any positive integer cannot be of the form $3m + 2$.

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Alternate Solution:

We can verify this by checking the squares of the first few positive integers and observing their remainders when divided by $3$:

Positive Integer ($n$)	Square ($n^2$)	Representation	Form
1	1	$3(0) + 1$	$3m + 1$
2	4	$3(1) + 1$	$3m + 1$
3	9	$3(3) + 0$	$3m$
4	16	$3(5) + 1$	$3m + 1$
5	25	$3(8) + 1$	$3m + 1$

Since the remainder is always either $0$ or $1$, the square of a positive integer can never leave a remainder of $2$ when divided by $3$. Therefore, it cannot be of the form $3m + 2$.

Given:

A positive integer is of the form $3q + 1$, where $q$ is a natural number.

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To Find:

Whether the square of this integer can be written in any form other than $3m + 1$, such as $3m$ or $3m + 2$, for some integer $m$.

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Solution:

Let the given positive integer be $n$. According to the question:

Now, we need to find the form of its square ($n^2$). Squaring both sides of equation (i):

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

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

From the first two terms, we can take $3$ as a common factor:

Since $q$ is a natural number, the expression $(3q^2 + 2q)$ will also be an integer. Let us denote this integer as $m$.

Substituting $m$ into equation (ii), we get:

This shows that the square of a number of the form $3q + 1$ is always of the form $3m + 1$. It cannot be of the form $3m$ or $3m + 2$.

Conclusion: No, we cannot write the square of a positive integer of the form $3q + 1$ in any form other than $3m + 1$.

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Alternate Solution:

We can justify this by taking different values for $q$ (natural numbers $1, 2, 3, \dots$) and checking the form of the square of $3q + 1$:

Value of $q$	Integer $n = 3q + 1$	Square $n^2$	Form $3m + r$
1	$3(1) + 1 = 4$	16	$3(5) + 1$
2	$3(2) + 1 = 7$	49	$3(16) + 1$
3	$3(3) + 1 = 10$	100	$3(33) + 1$
4	$3(4) + 1 = 13$	169	$3(56) + 1$

From the table above, we can observe that in every case, the square of the integer results in a remainder of $1$ when divided by $3$. This confirms that the form is always $3m + 1$ and never $3m$ or $3m + 2$.

Given:

The numbers are $525$ and $3000$.

The common divisors of these two numbers are $3, 5, 15, 25$ and $75$.

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To Find:

The Highest Common Factor, i.e., $\text{HCF} (525, 3000)$.

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Solution:

The Highest Common Factor (HCF) of two or more numbers is defined as the largest or greatest number among all the common factors (divisors) of those numbers.

From the given information, we list all the common divisors of $525$ and $3000$ in the table below:

Common Divisors	Values
Factor 1	3
Factor 2	5
Factor 3	15
Factor 4	25
Factor 5	75

By comparing these common divisors, we identify the greatest value:

Since $75$ is the largest among the common divisors, the HCF is $75$.

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Justification (Alternate Solution):

We can justify the answer using the prime factorisation method. Let us find the prime factors of both numbers:

Prime Factorisation of 525:

$\begin{array}{c|cc} 3 & 525 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

Prime Factorisation of 3000:

$\begin{array}{c|cc} 2 & 3000 \\ \hline 2 & 1500 \\ \hline 2 & 750 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The HCF is the product of the lowest powers of the common prime factors:

Common prime factors are $3$ and $5$.

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

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

$\text{HCF} = 75$

This confirms that the Highest Common Factor of $525$ and $3000$ is indeed $75$.

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.

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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.

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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.

Given:

The rational number is $\frac{987}{10500}$.

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To Find:

Whether the decimal expansion of $\frac{987}{10500}$ is terminating or non-terminating repeating, without using long division.

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Solution:

A rational number $\frac{p}{q}$ has a terminating decimal expansion if, when reduced to its simplest form (where $p$ and $q$ are co-prime), the prime factorisation of the denominator $q$ is of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers.

Step 1: Simplify the fraction to its lowest terms.

We check for common factors between $987$ and $10500$. Both numbers are divisible by $3$ (since the sum of digits of $987$ is $24$ and for $10500$ is $6$).

$\frac{987}{10500} = \frac{\cancel{987}^{329}}{\cancel{10500}_{3500}}$

Now, we check if $329$ and $3500$ have more common factors. $329$ is divisible by $7$ ($7 \times 47 = 329$) and $3500$ is also divisible by $7$.

$\frac{329}{3500} = \frac{\cancel{329}^{47}}{\cancel{3500}_{500}}$

Here, $47$ is a prime number, so $47$ and $500$ are co-prime.

Step 2: Prime factorisation of the denominator $q = 500$.

$\begin{array}{c|cc} 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorisation of the denominator is:

Since the denominator $500$ can be expressed in the form $2^n \times 5^m$ (where $n = 2$ and $m = 3$), the rational number has a terminating decimal expansion.

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Alternate Solution:

We can convert the fraction to a decimal by making the denominator a power of $10$ to prove it terminates.

From equation (i), we have $\frac{47}{500}$. To make the denominator a power of $10$, we multiply both numerator and denominator by $2$:

$\frac{47 \times 2}{500 \times 2} = \frac{94}{1000}$

Since the division results in $0.094$, it is a terminating decimal. The decimal terminates after three decimal places.

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Reasons for the answer:

1. The denominator of the simplified fraction contains only prime factors 2 and 5.

2. Any rational number whose denominator is of the form $2^n \times 5^m$ always results in a terminating decimal as per the Rational Number Theorem.

Given:

A rational number in its decimal expansion: $327.7081$.

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To Find:

The nature of the prime factors of the denominator $q$, when the number is expressed in the form $\frac{p}{q}$.

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Solution:

The given decimal number is $327.7081$. Since the digits after the decimal point stop (end) after four places, this is a terminating decimal expansion.

According to the theorem on rational numbers, if a rational number $\frac{p}{q}$ (where $p$ and $q$ are co-prime) has a terminating decimal expansion, then the prime factorisation of the denominator $q$ is of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers.

Let us express the given number in $\frac{p}{q}$ form:

Here, the denominator $q$ is $10000$. Let us find its prime factorisation:

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

We can see that the prime factors of $q$ consist only of the primes $2$ and $5$.

Reasons:

1. The decimal expansion is terminating, which implies the denominator must only have powers of $2$ and $5$.

2. If any other prime factor (like $3$ or $7$) were present in the denominator (and not cancelled by the numerator), the decimal expansion would have been non-terminating repeating.

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Alternate Solution:

We can directly apply the property of terminating decimals. Any terminating decimal can be written as a fraction where the denominator is a power of $10$.

Since $10 = 2 \times 5$, any power of $10$ will have only $2$ and $5$ as its prime factors.

Thus, we can conclude that the prime factors of $q$ are $2$ and $5$ only.

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Summary Table:

Decimal Form	Nature	Denominator ($q$)	Prime Factors of $q$
$327.7081$	Terminating	$10000$	$2, 5$

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.

Given:

Let $n$ be an odd positive integer.

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To Prove:

The square of $n$, i.e., $n^2$, is of the form $8m + 1$, where $m$ is some whole number.

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Proof:

According to Euclid’s Division Lemma, any positive integer can be expressed in the form $4q + r$, where $r = 0, 1, 2, 3$.

Therefore, any positive integer can be of the form $4q, 4q + 1, 4q + 2,$ or $4q + 3$.

Out of these, odd positive integers are of the form:

Now, let us consider the square of these odd integers in two cases:

Case I: When $n = 4q + 1$

Squaring both sides:

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

Let $m = 2q^2 + q$. Since $q$ is an integer, $m$ is a whole number.

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Case II: When $n = 4q + 3$

Squaring both sides:

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

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

To take $8$ as a common factor, we split $9$ into $8 + 1$:

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

Let $m = 2q^2 + 3q + 1$. Since $q$ is an integer, $m$ is a whole number.

From equations (i) and (ii), we conclude that the square of any odd positive integer is of the form $8m + 1$.

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Alternate Solution:

Any odd positive integer $n$ can be represented as $2k + 1$.

Squaring both sides:

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

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

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

We know that $k(k + 1)$ is the product of two consecutive integers, which is always even. Therefore, we can write $k(k + 1) = 2m$.

Substituting this in the expression:

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

$n^2 = 8m + 1$

Hence, the square of an odd positive integer is always of the form $8m + 1$.

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Verification Table:

Odd Integer ($n$)	Square ($n^2$)	Form ($8m + 1$)	Value of $m$
1	1	$8(0) + 1$	0
3	9	$8(1) + 1$	1
5	25	$8(3) + 1$	3
7	49	$8(6) + 1$	6

Given:

The expression is $\sqrt{2} + \sqrt{3}$.

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To Prove:

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

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Proof:

Let us assume, to the contrary, that $\sqrt{2} + \sqrt{3}$ is a rational number.

Therefore, we can write:

Rearranging the terms, we get:

$\sqrt{3} = a - \sqrt{2}$

Squaring both sides of the equation:

$(\sqrt{3})^2 = (a - \sqrt{2})^2$

Rearranging the terms to isolate the irrational part:

$2a\sqrt{2} = a^2 + 2 - 3$

$2a\sqrt{2} = a^2 - 1$

Since $a$ is a rational number, it follows that $a^2$, $a^2 - 1$, and $2a$ are also rational numbers. Thus, the entire expression $\frac{a^2 - 1}{2a}$ is a rational number.

From equation (i), this implies that $\sqrt{2}$ is a rational number.

However, this contradicts the well-known fact that $\sqrt{2}$ is an irrational number.

This contradiction has arisen because of our incorrect assumption that $\sqrt{2} + \sqrt{3}$ is rational.

Hence, we conclude that $\sqrt{2} + \sqrt{3}$ is irrational.

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Alternate Solution:

Let $x = \sqrt{2} + \sqrt{3}$. Squaring both sides directly:

$x^2 = (\sqrt{2} + \sqrt{3})^2$

$x^2 = (\sqrt{2})^2 + (\sqrt{3})^2 + 2(\sqrt{2})(\sqrt{3})$

$x^2 = 5 + 2\sqrt{6}$

$\sqrt{6} = \frac{x^2 - 5}{2}$

If $x$ is assumed to be rational, then $\frac{x^2 - 5}{2}$ must also be rational. This would mean $\sqrt{6}$ is rational.

But we know that the square root of any prime or product of distinct primes (like $6 = 2 \times 3$) is irrational. Thus, $\sqrt{6}$ is irrational.

Since a rational number cannot be equal to an irrational number, our assumption is false. Therefore, $\sqrt{2} + \sqrt{3}$ is irrational.

To Prove:

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. By Euclid's Division Lemma, if we take $b = 4$, then $a$ can be written in the form:

Where $r$ is an integer such that $0 \le r < 4$.

Therefore, the possible values of $r$ are $0, 1, 2,$ and $3$.

This implies that $a$ can be of the form $4m$, $4m + 1$, $4m + 2$, or $4m + 3$.

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Case I: When $a = 4m$

Squaring both sides:

$a^2 = (4m)^2$

$a^2 = 16m^2$

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Case II: When $a = 4m + 1$

Squaring both sides:

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

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

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Case III: When $a = 4m + 2$

Squaring both sides:

$a^2 = (4m + 2)^2$

$a^2 = 16m^2 + 16m + 4$

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Case IV: When $a = 4m + 3$

Squaring both sides:

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

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

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

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Conclusion:

Thus, in all the cases, we see that the square of any positive integer is either of the form $4q$ or $4q + 1$ for some integer $q$.

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Alternate Solution:

Any positive integer $n$ can be either even or odd.

Case 1: If $n$ is even

Let $n = 2k$ for some integer $k$.

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

Case 2: If $n$ is odd

Let $n = 2k + 1$ for some integer $k$.

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

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

Hence, the square of any positive integer is of the form $4q$ or $4q + 1$.

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

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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 $0 \leq r < 4$.

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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 $m$.

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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 $q$.

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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 $b=5$, there exist unique integers $k$ and $r$ such that $a = 5k + r$, where the remainder $r$ must satisfy the condition $0 \leq r < 5$.

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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 $k \geq 0$:

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

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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.

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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.

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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.

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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.

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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.

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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.

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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 two odd positive integers.

---

To Prove:

$(x^2 + y^2)$ is an even number but it is not divisible by 4.

---

Proof:

We know that any odd positive integer is of the form $2m + 1$ for some integer $m$.

Let $x = 2m + 1$ and $y = 2n + 1$ for some integers $m$ and $n$.

Now, consider the expression $x^2 + y^2$:

$x^2 + y^2 = (2m + 1)^2 + (2n + 1)^2$

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, we get:

$x^2 + y^2 = (4m^2 + 4m + 1) + (4n^2 + 4n + 1)$

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

Taking 2 common from the entire expression:

Since $x^2 + y^2$ is a multiple of 2, it is an even number.

Now, let us rewrite the expression to check for divisibility by 4:

$x^2 + y^2 = 4(m^2 + n^2 + m + n) + 2$

Let $k = m^2 + n^2 + m + n$, where $k$ is an integer.

The above equation is in the form of $Dividend = Divisor \times Quotient + Remainder$.

Here, the remainder is 2 when $x^2 + y^2$ is divided by 4.

Since the remainder is not zero, $x^2 + y^2$ is not divisible by 4.

Hence, if $x$ and $y$ are odd positive integers, then $x^2 + y^2$ is even but not divisible by 4.

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.

Given:

The numbers are 1251, 9377, and 15628.

The remainders left after division are 1, 2, and 3 respectively.

---

To Find:

The largest number (HCF) that divides these numbers leaving the specified remainders.

---

Solution:

The largest number that divides 1251, 9377, and 15628 leaving remainders 1, 2, and 3 respectively must be the Highest Common Factor (HCF) of the numbers obtained after subtracting the remainders.

The required numbers are:

$1251 - 1 = 1250$

$9377 - 2 = 9375$

$15628 - 3 = 15625$

Now, we need to find the HCF of 1250, 9375, and 15625 using Euclid's division algorithm.

---

Step 1: Finding HCF of 1250 and 9375.

Since $9375 > 1250$, we apply Euclid's division lemma to 9375 and 1250:

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

Since the remainder is now 0, the divisor at this stage is the HCF.

Therefore, HCF(1250, 9375) = 625.

---

Step 2: Finding HCF of 625 and 15625.

Now, we apply Euclid's division lemma to the third number 15625 and the HCF obtained in Step 1 (625):

Since the remainder is 0, the HCF of 1250, 9375, and 15625 is 625.

---

Final Answer:

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

---

Alternate Solution:

We can also verify the HCF using prime factorisation (though the question specifically asks for Euclid's algorithm):

Prime factorisation of 1250:

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

$1250 = 2 \times 5^4$

Prime factorisation of 9375:

$\begin{array}{c|cc} 3 & 9375 \\ \hline 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$9375 = 3 \times 5^5$

Prime factorisation of 15625:

$\begin{array}{c|cc} 5 & 15625 \\ \hline 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$15625 = 5^6$

Common factor = $5^4 = 625$.

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:

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

---

Proof:

Let us consider the prime factorisation of the number 12.

The prime factorisation of 12 is as follows:

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

So, we can write:

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

Now, for any natural number $n$, the prime factorisation of $12^n$ will be:

---

According to the Fundamental Theorem of Arithmetic, the prime factorisation of every composite number is unique, apart from the order in which the prime factors occur.

From equation (i), it is evident that the only prime factors of $12^n$ are 2 and 3.

---

Condition for a number to end with the digit 0:

For any positive integer to end with the digit 0, it must be divisible by 10. This means its prime factorisation must contain both 2 and 5 as prime factors.

In the case of $12^n$, the prime factorisation contains 2, but it does not contain the prime factor 5.

Hence, $12^n$ cannot end with the digit 0.

---

Condition for a number to end with the digit 5:

For any positive integer to end with the digit 5, it must be divisible by 5. This means its prime factorisation must contain the prime factor 5.

Furthermore, if it ends in 5, it must be an odd number, meaning it should not contain 2 as a factor.

In the case of $12^n$, the prime factorisation does not contain 5, and it does contain 2.

Hence, $12^n$ cannot end with the digit 5.

---

Conclusion:

Since the prime factorisation of $12^n$ does not contain the prime factor 5, it is impossible for $12^n$ to end with the digit 0 or 5 for any natural number $n$.

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

---

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

---

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 $r$ is $0, 1, 2, 3, 4,$ or $5$, is also of the form $6m + r$ for some integer $m$.

---

Proof:

Let $n$ be a positive integer such that:

Here, $r$ can be any integer from the values $0, 1, 2, 3, 4, 5$.

Now, we need to find the cube of $n$, denoted as $n^3$.

Taking the cube on both sides of equation (i):

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

Using the algebraic identity $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, we expand the right-hand side:

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

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

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

Factoring out 6 from the first three terms, we get:

So, the expression becomes:

Now, we will substitute each possible value of $r$ into equation (ii).

---

Case 1: When $r = 0$

$n^3 = 6K + (0)^3 = 6K$

---

Case 2: When $r = 1$

$n^3 = 6K + (1)^3 = 6K + 1$

---

Case 3: When $r = 2$

$n^3 = 6K + (2)^3 = 6K + 8$

$n^3 = 6K + 6 + 2$

---

Case 4: When $r = 3$

$n^3 = 6K + (3)^3 = 6K + 27$

$n^3 = 6K + 24 + 3$

---

Case 5: When $r = 4$

$n^3 = 6K + (4)^3 = 6K + 64$

$n^3 = 6K + 60 + 4$

---

Case 6: When $r = 5$

$n^3 = 6K + (5)^3 = 6K + 125$

$n^3 = 6K + 120 + 5$

---

Conclusion:

From all the cases above, we observe that for every possible value of $r$, the cube of the integer $n$ remains in the form $6m + r$.

Hence, it is proved that the cube of a positive integer of the form $6q + r$ is also of the form $6m + r$.

To Prove:

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

---

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.