What will be remainder when pow(23,80) is divided by 24 ?

17

16

1

2

When n is even. (x^n - a^n) is completely divisible by (x + a) (23^80 - 1^80) is completely divisible by (23 + 1), i.e., 24. (23^80 - 1) is completely divisible by 24. On dividing 23^80 by 24, we get 1 as remainder.

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