Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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(Work in progress of my answer to this question.) |
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<math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math> | <math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math> | ||
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| + | 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 + 63 = 20x + 12y</cmath>, | ||
| + | <cmath>x\cdot y - 20x - 12y + 63 = 0</cmath>. | ||
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| + | (awesomeag) | ||
Revision as of 16:22, 16 February 2018
23. 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?
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)
.
(awesomeag)
