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 06:05, 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: ![\[110\text{ some number of zeros }011.\]](http://latex.artofproblemsolving.com/8/8/d/88d5c619fae9d43d5d3ac869d08832df3644c2c4.png)
This works because of the way they are multiplied. Therefore, the answer is 
.
