Difference between revisions of "2018 AMC 10B Problems/Problem 23"
| Line 9: | Line 9: | ||
==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 |
<cmath>x\cdot y + 63 = 20x + 12y</cmath> | <cmath>x\cdot y + 63 = 20x + 12y</cmath> | ||
Revision as of 19:41, 15 January 2021
Contents
Problem
How many ordered pairs 
of positive integers satisfy the equation
![\[a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),\]](http://latex.artofproblemsolving.com/1/8/d/18dfc63240b01569b535ba05a0d53841ad6cfe8b.png)
where 
denotes the greatest common divisor of 
and 
, and 
denotes their least common multiple?
Solution
Let $x = \lcm(a, b)$ (Error compiling LaTeX. Unknown error_msg), and 
. Therefore, $a\cdot b = \lcm(a, b)\cdot \gcd(a, b) = x\cdot y$ (Error compiling LaTeX. Unknown error_msg). Thus, the equation becomes
![\[x\cdot y + 63 = 20x + 12y\]](http://latex.artofproblemsolving.com/f/a/b/fabf3251ab45c9a46352eae222bfb87de8a20b11.png)
![\[x\cdot y - 20x - 12y + 63 = 0\]](http://latex.artofproblemsolving.com/8/2/b/82bc19a6185adaf76a5daddc460ab02f14038347.png)
Using Simon's Favorite Factoring Trick, we rewrite this equation as
![\[(x - 12)(y - 20) - 240 + 63 = 0\]](http://latex.artofproblemsolving.com/f/1/5/f15f9030f3a91e97f5177a080c2c67d39bb18996.png)
![\[(x - 12)(y - 20) = 177\]](http://latex.artofproblemsolving.com/5/f/a/5fa4468ebc4559a0d0b039d0184912c1b30586cc.png)
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. 
