# Difference between revisions of "1975 AHSME Problems/Problem 10"

## Problem

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

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4n \qquad \textbf{(C)}\ 2+2n \qquad \textbf{(D)}\ 4n^2 \qquad \textbf{(E)}\ n^2+n+2$

## Solution

We see that the result of this expression will always be in the form $(100\text{ some number of zeros }001)^2.$ Multiplying these together yields: $$110\text{ some number of zeros }011.$$ This works because of the way they are multiplied. Therefore, the answer is $\boxed{(A) 4}$.