Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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− | + | ==Problem== | |
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+ | 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> | ||
where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple? | where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple? | ||
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− | ==Solution | + | ==Solution== |
Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes | Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes | ||
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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]], we rewrite this equation as |
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath> | <cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath> | ||
<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 <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. |
(awesomeag) | (awesomeag) | ||
+ | |||
+ | Edited by IronicNinja, Firebolt360, and mprincess0229~ | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=JWGHYUeOx-k | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Revision as of 12:35, 14 April 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 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
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.