Given below are two statements :Statement I :25^13 + 20^13 + 8^13 + 3^13 is divisible by 7…

Question

Given below are two statements :Statement I :$25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by 7.Statement II :The integral part of $(7 + 4\sqrt{3})^{25}$ is an odd number.In the light of the above statements, choose the correct answer ...

Accepted Answer

Statement I : $25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by 7.

Key concept : If $n$ is odd, $x^n+y^n=(x+y)\left(x^{n-1}-x^{n-2} y+x^{n-3} y^2-\cdots-x y^{n-2}+\right. y^{n-1}$ ) So, $x^n+y^n$ is divisible by $x+y$.

In the given statement :

$25^{13}+3^{13}$ is divisible by $25+3=28$.

$25^{13}+3^{13}$ is also divisible by 7 because 28 is divisible by 7 .

So, $25^{13}+3^{13}=7 \cdot I_1$.........(1) ( $I_1$ is a positive integer).

Similarly :

$20^{13}+8^{13}$ is divisible by $20+8=28$.

$20^{13}+8^{13}$ is divisible by 7.

So, $20^{13}+8^{13}=7 \cdot I_2$ ..........(2) ( $I_2$ is a positive integer).

Add (1) \& (2) :

$25^{13}+3^{13}+20^{13}+8^{13}=7\left(I_1+I_2\right)$

So, $25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by 7 . Statement I is true.

Statement : The integral part of $(7+4 \sqrt{3})^{25}$ is an odd number.

Let $X=(7+4 \sqrt{3})^{25}$. Let $I$ be the integral part and $f$ be the fractional part ( $0Adding the two expansions:

$$ (7+4 \sqrt{3})^{25}+(7-4 \sqrt{3})^{25}=I+f+f^{\prime} $$

Using the binomial expansion $(a+b)\^n+(a-b)\^n=2\left[a\^n+\binom{n}{2} a\^{n-2} b\^2+\ldots\right]$, we see that the sum is always an even integer because of the factor of 2 .

$$ I+f+f^{\prime}=\text { Even Integer } $$

Since $I$ is an integer, $f+f\^{\prime}$ must also be an integer. Given $0 $$ I+1=\text { Even Integer } $$

$$ I=\text { Even Integer }-1=\text { Odd Integer } $$

The integral part I is indeed an odd number. Statement II is True.

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