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1. The remainder of $\frac{39^{97!}}{40}$ is :

A. 39

B. 0

C. 1

D. 13

Solution:

Since,

\frac{a^{n}}{a + 1}

gives remainder 1 when 'n' is even.
Now since, 97! is an even number so remainder will be 1.
Note:-Factorial of any no. is even.

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