Difference between revisions of "1975 AHSME Problems/Problem 10"
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− | + | ==Problem== | |
− | ==Problem | ||
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== | ||
+ | {{AHSME box|year=1975|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:54, 19 January 2021
Problem
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 .
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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