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

A. 39

B. 0

C. 1

D. 13

E. None of these

Answer: Option C

Solution(By Examveda Team)

Since, $$\frac{{{{\text{a}}^{\text{n}}}}}{{{\text{a}} + 1}}$$   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.