# 2018 AMC 10B Problems/Problem 23

23. How many ordered pairs $(a, b)$ of positive integers satisfy the equation $$a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),$$ where $\text{gcd}(a,b)$ denotes the greatest common divisor of $a$ and $b$, and $\text{lcm}(a,b)$ denotes their least common multiple?

$\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}$

## Solution 1

Let $x = lcm(a, b)$, and $y = gcd(a, b)$. Therefore, $a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y$. Thus, the equation becomes

$$x\cdot y + 63 = 20x + 12y$$ $$x\cdot y - 20x - 12y + 63 = 0$$

Using Simon's Favorite Factoring Trick, we rewrite this equation as

$$(x - 12)(y - 20) - 240 + 63 = 0$$ $$(x - 12)(y - 20) = 177$$

Since $177 = 3\cdot 59$ and $x > y$, we have $x - 12 = 59$ and $y - 20 = 3$, or $x - 12 = 177$ and $y - 20 = 1$. This gives us the solutions $(71, 23)$ and $(189, 21)$, which can be translated back to two solution for $a$ and $b$. Thus, the answer is $\boxed{2}$. (awesomeag)