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When $32^{\left(32^{32)}\right.}$ is divided by $7,$ the remainder is ______ .
Option: 1
$2$

Option: 2
$3$

Option: 3
$4$

Option: 4
$9$

Answers (1)

We have $32^{32}=\left(2^5\right)^{32}=2^{160}$

$=(3-1)^{160}={ }^{160} C_0 3^{160}-{ }^{160} C_1 3^{159}+\ldots-{ }^{160} C_{159} \cdot 3+1$

$=3 m+1,$ where $m$ is some positive integer.
Now, $32^{\left(32^{32)}\right.}=\left(2^5\right)^{3 m+1}=\left(2^3\right)^{5 m+1} \cdot 2^2=(7+1)^{5 m+1} \cdot 4$

But $(7+1)^{5 m+1}=7 n+1$ for some positive integer $n$ . Thus, $N=32^{\left(32^{32)}\right.}=28 n+4 .$

Thus shows that when $N$ is divided by $7,$ the remainder is 4 $4.$

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Kshitij

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