Question

The sum of digits of the number ${({10}^{4n^2+8}+1)}^2$, where $n$ is a positive integer, is

• A. $4$
• B. $8$
• C. $9$
• D. $11$

Answer 1

The correct option is A $4$

When we put any positive value of $n$, we get answer in such a way that we get two one at the start and one at the end and in the middle we get two.
So the sum will always be $4$.