\( \Large 19^{5}+21^{5} \) is divisible by


A) Only 10

B) Only 20

C) Both 10 and 20

D) Neither 10 nor 20

Correct Answer:
C) Both 10 and 20

Description for Correct answer:

We can check divisibility of \( \Large 19^{5}+21^{5} \) by

10 by adding the unit digits of \( \Large 9^{5} \  and \  1^{5} \) which is equal to 9 + 1 = 10.

So, it must be divisible by 10.

Now, for divisibility by 20 we add 19 and

21 which is equal to 40. So, it is clear that it is also divisible by 20.

So, \( \Large 19^{5}+21^{5} \) is divisible by both 10 and 20.


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