The remainder of $$\frac{{{{39}^{97!}}}}{{40}}$$ is :
A. 39
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.
This Question Belongs to Arithmetic Ability >> Number System
the ans option is "c"