Difference between revisions of "2020 AIME II Problems/Problem 1"
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==Solution== | ==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 | + | 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 | ||
+ | <br><br> | ||
+ | ==Solution 2 (Official MAA)== | ||
+ | Because <math>20^{20}=2^{40}5^{20}</math>, if <math>m^2n = 20^{20}</math>, there must be nonnegative integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> such that | ||
+ | <math>m = 2^a5^b</math> and <math>n = 2^c5^d</math>. Then | ||
+ | <cmath>2a + c = 40</cmath> | ||
+ | and | ||
+ | <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>. | ||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zfChnbMGLVQ?t=4612 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=x0QznvXcwHY | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution == | ||
+ | |||
+ | https://youtu.be/Va3MPyAULdU | ||
+ | |||
+ | ~avn | ||
+ | |||
+ | ==Purple Comet Math Meet April 2020== | ||
+ | |||
+ | Notice, that this was the exact same problem (with different wording of course) as Purple Comet HS problem 3 and remembering the answer, put <math>\boxed{231}</math>. | ||
+ | |||
+ | https://purplecomet.org/views/data/2020HSSolutions.pdf | ||
+ | |||
+ | ~Lopkiloinm | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2020|n=II|before=First Problem|num-a=2}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:27, 17 January 2021
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://youtu.be/zfChnbMGLVQ?t=4612
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=x0QznvXcwHY
~IceMatrix
Video Solution
~avn
Purple Comet Math Meet April 2020
Notice, that this was the exact same problem (with different wording of course) as Purple Comet HS problem 3 and remembering the answer, put .
https://purplecomet.org/views/data/2020HSSolutions.pdf
~Lopkiloinm
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.