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Difference between revisions of "1975 AHSME Problems/Problem 10"
 
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==Problem==
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==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
 
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
  
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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> This works because of the way they are multiplied. Therefore, the answer is <math>\boxed{(A) 4}</math>.
 
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>.
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==See Also==
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{{AHSME box|year=1975|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 15:54, 19 January 2021

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 by
Problem 9 		Followed by
Problem 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

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