If 7 divides ${32^{32}}^{32}$ , the remainder is

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Solution

The correct option is B 4
$32 = 2^{5} \Rightarrow (32)^{32} = (2^{5})^{32}$
( $2^{160} = (3 - 1)^{160} = 3 m + 1, m \in N$
$∴ {(32)^{32}}^{32} = (32) 3 m + 1 = 2^{5} (3 m + 1)$
$2^{3 (5 m + 1)} 2^{2} = {4.8}^{5 m + 1}$
$(7 + 1)^{5 m + 1} = 4 (7 n + 1), n \in N = 28 n + 4$
$∴$ $W h e n 7 d i v i d e s$ ${(32)^{32}}^{32}$ $r e m a i n d e r = 4$

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