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>. This works because of the way they are multiplied. Therefore, the answer is <math>\boxed{(A) 4}</math>.
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==Problem==
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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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<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>
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==Solution==
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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>.
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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 16:54, 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 (ProblemsAnswer KeyResources)
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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