Difference between revisions of "2020 AIME II Problems/Problem 1"
m (→Solution 2 (Official MAA)) |
m |
||
Line 13: | Line 13: | ||
<cmath>2a + c = 40</cmath> | <cmath>2a + c = 40</cmath> | ||
and | and | ||
− | <cmath>2b+d | + | <cmath>2b+d = 20</cmath> |
The first equation has <math>21</math> solutions corresponding to <math>a = 0,1,2,\dots,20</math>, and the second equation has <math>11</math> solutions corresponding to <math>b = 0,1,2,\dots,10</math>. Therefore there are a total of <math>21\cdot11 = 231</math> ordered pairs <math>(m,n)</math> such that <math>m^2n = 20^{20}</math>. | The first equation has <math>21</math> solutions corresponding to <math>a = 0,1,2,\dots,20</math>, and the second equation has <math>11</math> solutions corresponding to <math>b = 0,1,2,\dots,10</math>. Therefore there are a total of <math>21\cdot11 = 231</math> ordered pairs <math>(m,n)</math> such that <math>m^2n = 20^{20}</math>. | ||
Revision as of 09:43, 8 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
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
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.