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The remainder of $$\frac{{{{39}^{97!}}}}{{40}}$$ is :
Answer & Solution
Correct Answer:
Option
C
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.
Now since, 97! is an even number so remainder will be 1.
Note:-Factorial of any no. is even.
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