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Jee Main 2024
29-01-2024 S2
Question
Remainder when $64^{32^{32}}$ is divided by 9 is equal to
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Solution
Let $32^{32}=\mathrm{t}$ $$ \begin{aligned} & 64^{32^{32}}=64^t=8^{2 t}=(9-1)^{2 t} \\ & =9 k+1 \end{aligned} $$ Hence remainder = 1
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