Difference between revisions of "2020 AIME II Problems/Problem 1"
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+ | ==Video Solution by WhyMath== | ||
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+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 07:20, 30 May 2023
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 by OmegaLearn
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
Video Solution by WhyMath
~savannahsolver
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.