Question

If $7$ divides ${32}^{{32}^{32}}$, then find the remainder.

Answer 1

${32}^{2^{32\times 5}}$ $=$ ${32}^{2^{160}}$ $=$ ${64}^{2^{159}}$ $=$ ${(63+1)}^k$

Let us suppose $2^{159}$ as $k$

${(63+1)}^k$ $=$ ${}^k{C}_0$.${63}^k$ $+$ ${}^k{C}_1$.${63}^{k-1}$..........+${}^k{C}_{k-1}$.$63$ $+$ ${}^k{C}_k$

${(63+1)}^k$ $=$ ${}^k{C}_0$.${63}^k$ $+$ ${}^k{C}_1$.${63}^{k-1}$..........+${}^k{C}_{k-1}$.$63$ $+1$

${(63+1)}^k$ $=63a+1$

${(63+1)}^k$ $=7A+1$

Hence, the remainder is $1$.