# 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 $\boxed{(A) 4}$.

## See Also

 1975 AHSME (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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