Difference between revisions of "2020 AIME II Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | + | In this problem, we want to find the number of ordered pairs <math>(m, n)</math> such that <math>m^2n = 20^{20}</math>. Let <math>x = m^2</math>. Therefore, we want two numbers, <math>x</math> and <math>n</math>, such that their product is <math>20^{20}</math> and <math>x</math> is a perfect square. Note that there is exactly one valid <math>n</math> for a unique <math>x</math>, which is <math>\tfrac{20^{20}}{x}</math>. This reduces the problem to finding the number of unique perfect square factors of <math>20^{20}</math>. | |
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+ | <math>20^{20} = 2^{40} \cdot 5^{20} = \left(2^2\right)^{20}\cdot\left(5^2\right)^{10}.</math> Therefore, the answer is <math>21 \cdot 11 = \boxed{231}.</math> | ||
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~superagh | ~superagh | ||
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+ | ~TheBeast5520 | ||
==Solution 2 (Official MAA)== | ==Solution 2 (Official MAA)== | ||
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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>. | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/zfChnbMGLVQ?t=4612 | ||
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+ | ~ pi_is_3.14 | ||
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+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=VA1lReSkGXU | ||
+ | |||
+ | ~ North America Math Contest Go Go Go | ||
==Video Solution== | ==Video Solution== | ||
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~IceMatrix | ~IceMatrix | ||
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+ | ==Video Solution == | ||
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+ | https://youtu.be/Va3MPyAULdU | ||
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+ | ~avn | ||
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+ | ==Purple Comet Math Meet April 2020== | ||
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+ | 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>. | ||
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+ | https://purplecomet.org/views/data/2020HSSolutions.pdf | ||
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+ | ~Lopkiloinm | ||
==See Also== | ==See Also== |
Latest revision as of 16:25, 7 March 2021
Contents
Problem
Find the number of ordered pairs of positive integers such that .
Solution
In this problem, we want to find the number of ordered pairs such that . Let . Therefore, we want two numbers, and , such that their product is and is a perfect square. Note that there is exactly one valid for a unique , which is . This reduces the problem to finding the number of unique perfect square factors of .
Therefore, the answer is
~superagh
~TheBeast5520
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=VA1lReSkGXU
~ North America Math Contest Go Go Go
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.