Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath> | <cmath>x\cdot y - 20x - 12y + 63 = 0</cmath> | ||
− | Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as | + | Using [[Simon's Favorite Factoring Trick]]or SFFT, we rewrite this equation as |
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath> | <cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath> |
Revision as of 16:35, 27 January 2020
Contents
Problem
How many ordered pairs of positive integers satisfy the equation where denotes the greatest common divisor of and , and denotes their least common multiple?
Solution
Let , and . Therefore, . Thus, the equation becomes
Using Simon's Favorite Factoring Trickor SFFT, we rewrite this equation as
From here we can already see that this is a quadratic, and thus must have solutions. But, let's continue, to see if one of the solutions is extraneous.
Since and , we have and , or and . This gives us the solutions and . Since the must be a divisor of the , the first pair does not work. Assume . We must have and , and we could then have , so there are solutions. (awesomeag)
Edited by IronicNinja, Firebolt360, and mprincess0229~
Video Solution
https://www.youtube.com/watch?v=JWGHYUeOx-k
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.