Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation | How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation | ||
<cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath> | <cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath> |
Revision as of 00:54, 5 December 2019
Contents
[hide]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 Trick, we rewrite this equation as
From here we can already see that this is a quadratic, and thus must have 2 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 GCD must be a divisor of the LCM, the first pair does not work. Assume . We must have and , and we could then have , so there are solutions. (awesomeag)
Edited by IronicNinja and Firebolt360~
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.