The sum of the digits in base ten of (10^4n^2+8+1)^2, where n is a positive integer, is

1975 AMC 12 Problem 10

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10.

The sum of the digits in base ten of

\left(10^{4n^2+8}+1\right)^2,

where

n

is a positive integer, is

4

4n

2+2n

4n^2

n^2+n+2

Set

k=4n^2+8

and square

10^k+1

The three nonzero digits occur in distinct places, so there are no carries

For

k=4n^2+8,

(10^k+1)^2=10^{2k}+2\cdot10^k+1.

Its only nonzero digits are

1,2,1,

so their sum is

4.

Therefore, the correct answer is A.