Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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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. | 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. | ||
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Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the Greatest Common Denominator must be a divisor of the Lowest Common Multiple, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions. | Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the Greatest Common Denominator must be a divisor of the Lowest Common Multiple, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions. | ||
(awesomeag) | (awesomeag) |
Revision as of 17:39, 11 February 2019
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 Greatest Common Denominator must be a divisor of the Lowest Common Multiple, 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.