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2 + 2 96 + 3 96 + 4 96 + + 9 6 96 \large 2+2^{96}+3^{96}+4^{96}+\ldots+96^{96}

Find the remainder when the above expression is divided by 97?


The answer is 0.

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1 solution

Siddharth Singh
Jun 20, 2015

This expression can be written as: \text{This expression can be written as:}

1 96 + 2 96 + 3 96 + 4 96 + . . . . . . . . . . . . + 9 6 96 + 1 1^{96}+2^{96}+3^{96}+4^{96}+............+96^{96}+1

And If p is prime then(can be proved by fermat theorem)

1 p 1 + 2 p 1 + 3 p 1 + . . . . . . . . . . . + ( p 1 ) p 1 + 1 0 ( m o d p ) 1^{p-1}+2^{p-1}+3^{p-1}+...........+(p-1)^{p-1}+1 \equiv 0 (mod p)

Since 97 is prime we can place 97 as the value of p.Therefore

1 97 1 + 2 97 1 + 3 97 1 + 4 97 1 + . . . . . . . . . . . . + 9 6 97 1 + 1 0 ( m o d 97 ) 1^{97-1}+2^{97-1}+3^{97-1}+4^{97-1}+............+96^{97-1}+1\equiv 0(mod 97)

Remainder= 0 \boxed{0}

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