Difference between revisions of "2020 AIME II Problems/Problem 1"
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− | First, we find the prime factorization of <math>20^{20}</math>, which is <math>2^{40}\times5^{20}</math>. The equation <math>{m^2n = 20 ^{20}}</math> tells us that we want to select a perfect square factor of <math>20^{20}</math>, <math>m^2</math>. The <math>n</math> might throw you off here, but it's actually kind of irrelevant because once <math>m</math> is selected, the remaining factor will already be assigned as <math>\frac{20^20}{m^2}</math>. There are <math>21\cdot11=231</math> ways to select a perfect square factor of <math>20^{20}</math>, thus our answer is <math>\boxed{231}</math>. | + | First, we find the prime factorization of <math>20^{20}</math>, which is <math>2^{40}\times5^{20}</math>. The equation <math>{m^2n = 20 ^{20}}</math> tells us that we want to select a perfect square factor of <math>20^{20}</math>, <math>m^2</math>. The <math>n</math> might throw you off here, but it's actually kind of irrelevant because once <math>m</math> is selected, the remaining factor will already be assigned as <math>\frac{20^{20}}{m^2}</math>. There are <math>21\cdot11=231</math> ways to select a perfect square factor of <math>20^{20}</math>, thus our answer is <math>\boxed{231}</math>. |
~superagh | ~superagh |
Revision as of 21:35, 24 September 2020
Contents
Problem
Find the number of ordered pairs of positive integers such that .
Solution
First, we find the prime factorization of , which is . The equation tells us that we want to select a perfect square factor of , . The might throw you off here, but it's actually kind of irrelevant because once is selected, the remaining factor will already be assigned as . There are ways to select a perfect square factor of , thus our answer is .
~superagh
Solution 2 (Official MAA)
Because , if , there must be nonnegative integers , , , and such that and . Then and The first equation has solutions corresponding to , and the second equation has solutions corresponding to . Therefore there are a total of ordered pairs such that .
Video Solution
https://www.youtube.com/watch?v=x0QznvXcwHY
~IceMatrix
Video Solution 2
~avn
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.