If ${32^{32}}^{32}$ is divided by 7 , then the remainder is

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Solution

The correct option is D
4

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

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

= 3 $(3^{159} -^{160} C_{1} 3^{158} + . . . . . -^{160} C_{159}) + 1$

= 3m + 1 , m $\in Z_{+}$

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

= $2^{3 (5 m + 1) {.2}^{2} = 4. (8)^{5 m + 1} = 4. (7 + 1)^{5 m + 1}}$

$= 4. [^{5 m + 1} C_{0} (7)^{5 m + 1} +^{5 m + 1} C_{1} (7)^{5 m} +^{5 m + 1} C_{2} (7)^{5 m - 1} + . . . +^{5 m + 1} C_{5 m} 7 +^{5 m + 1} C_{5 m + 1}$ ]

= $4 [7 n + 1], n \in X_{+}$ = 28n + 4

This shows that when ${32^{32}}^{32}$ is divisible by 7 , the remainder is 4.

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