# 2018 AMC 10B Problems/Problem 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

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

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

From here we can already see that this is a quadratic, and thus must have 2 solutions. But, let's continue, to see if one of the solutions is extraneous.

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 $(189, 21)$. Since the Greatest Common Denominator must be a divisor of the Lowest Common Multiple, the first pair does not work. Assume $a>b$. We must have $a = 21 \cdot 9$ and $b = 21$, and we could then have $a, so there are $\boxed{2}$ solutions. (awesomeag)

Edited by IronicNinja and Firebolt360~

## Video Solution

 2018 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 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