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>. Obviously, the first pair does not work. | + | 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>. Obviously, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 8</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 14:49, 23 September 2018
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 
, and 
. Therefore, 
. 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 
. Obviously, the first pair does not work. Assume 
. We must have 
and 
, and we could then have 
, so there are 
solutions. .
(awesomeag)
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. 
