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1. The remainder of $\frac{32^{32^{32}}}{7} :$

A. 1

B. 2

C. 3

D. 4

Solution:

\begin{aligned} 32^{32^{32}} means 32^{(32.32.32.32.......32 times)} \\ and the remainder of \frac{32}{7} is 4 \\ So, \\ \frac{4^{(32.32.32.32.......32 times)}}{7} \\ \frac{4^{(2.2.2.2.2........32 times)}}{7} \\ Remainder = 4 \\ Since, \frac{4}{7} \to Remainder 4 \\ \frac{4^{2}}{7} \to Remainder 2 \\ \frac{4^{3}}{7} \to Remainder 1 \\ \frac{4^{4}}{7} \to Remainder 4 \end{aligned}

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