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If 7 divides $3 2^{3 2^{32}}$ , the remainder is

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$32 = 2^{5} \Rightarrow (32)^{32} = (2^{5})^{32}$ $= 2^{160} = (3 - 1)^{160} = 3 m + 1, m \in N$ $∴ (32)^{3 2^{32}} = (32)^{3 m + 1} = 2^{5 (3 m + 1)}$ $2^{3 (5 m + 1)} 2^{2} = 4. 8^{5 m + 1}$ $4 (7 + 1)^{5 m + 1} = 4 (7 n + 1), n \in N = 28 n + 4$ ∴ When 7 divides $(32)^{3 2^{32}}$ remainder $= 4$

Stream:JEESubject:MathematicsTopic:Binomial Theorem

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