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

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− | We see that the result of this expression will always be in the form <math>(100\text{ some number of zeros}001)^2.</math> Multiplying these together yields: <cmath>110\text{ some number of zeros}011</cmath> | + | |

+ | ==Problem 10== | ||

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

+ | |||

+ | <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4n \qquad \textbf{(C)}\ 2+2n \qquad \textbf{(D)}\ 4n^2 \qquad \textbf{(E)}\ n^2+n+2</math> | ||

+ | ==Solution== | ||

+ | |||

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

## Revision as of 07:05, 17 May 2016

## Problem 10

The sum of the digits in base ten of , where is a positive integer, is

## Solution

We see that the result of this expression will always be in the form Multiplying these together yields: This works because of the way they are multiplied. Therefore, the answer is .