When ${(32)}^{{(32)}^{32}}$ is divided by $7$ , then the remainder is

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Solution

$({(32)}^{32}) = {(2^{5})}^{32} = {(2)}^{160} = {(3 - 1)}^{160}$

$=^{160} C_{0} {(3)}^{160} -^{160} C_{1} {(3)}^{159} + . . +^{160} C_{159} (3) +^{160} C_{160} {(3)}^{0}$

$= 3 k + 1,$ where $k \in N$

Now,

${32^{(32)}}^{32} = 32^{(3 k + 1)} = {(2^{5})}^{(3 k + 1)} = 2^{(15 k + 5)}$

$= 2^{3 (5 k + 1)} \times 2^{2} = 2^{3 (5 k + 1)} \times 4 = 4 {(7 + 1)}^{5 k + 1}$

$= 4 [^{5 k + 1} C_{0} 7^{5 k + 1} +^{5 k + 1} C_{1} 7^{5 k} + \dots +^{5 k + 1} C_{5 k} 7 +^{5 k + 1} C_{5 k + 1} \times 7 + 1]$

$= 4 (7 n + 1),$ where $n \in N$

$= 28 n + 4$

Therefore, when ${32^{(32)}}^{32}$ is divided by $7$ , the remainder is $4$ .

Hence, the answer is $4$ .

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