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>. <math>n</math> will be assigned by default. There are <math>21\times11=231</math> ways to select a perfect square factor of <math>20^{20}</math>, thus our answer is <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>. <math>n</math> will be assigned by default. There are <math>21\times11=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 20:56, 7 June 2020
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 , 
. 
will be assigned by default. There are 
ways to select a perfect square factor of , thus our answer is 
.
~superagh
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
