Difference between revisions of "2020 AIME II Problems/Problem 1"
Advancedjus (talk | contribs) |
(→Problem) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
Find the number of ordered pairs of positive integers <math>(m,n)</math> such that <math>{m^2n = 20 ^{20}}</math>. | Find the number of ordered pairs of positive integers <math>(m,n)</math> such that <math>{m^2n = 20 ^{20}}</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | 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>\mbox{231}</math>. | ||
+ | ~superagh |
Revision as of 17:39, 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