Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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<cmath>(x - 12)(y - 20) = 177</cmath> | <cmath>(x - 12)(y - 20) = 177</cmath> | ||
− | 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 <math>\text{GCD}</math> must be a divisor of the <math>\text{LCM}</math>, 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 <math>\text{GCD}</math> must be a divisor of the <math>\text{LCM}</math>, 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{\textbf{(B)} ~2}</math> solutions. |
(awesomeag) | (awesomeag) | ||
Edited by IronicNinja, Firebolt360, and mprincess0229~ | Edited by IronicNinja, Firebolt360, and mprincess0229~ | ||
− | == Video Solution == | + | == Video Solution by OmegaLearn== |
https://youtu.be/zfChnbMGLVQ?t=494 | https://youtu.be/zfChnbMGLVQ?t=494 | ||
Latest revision as of 04:18, 21 January 2023
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
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 by OmegaLearn
https://youtu.be/zfChnbMGLVQ?t=494
~ pi_is_3.14
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.