2018 AMC 10B Problems/Problem 23

Problem

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 $x = \text{lcm}(a, b)$ , and $y = \text{gcd}(a, b)$ . 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 $(189, 21)$ . Since the $\text{GCD}$ must be a divisor of the $\text{LCM}$ , the first pair does not work. Assume $a>b$ . We must have $a = 21 \cdot 9$ and $b = 21$ , and we could then have $a<b$ , so there are $\boxed{\textbf{(C)}2}$ solutions. (awesomeag)

Edited by IronicNinja, Firebolt360, and mprincess0229~

Video Solution

https://youtu.be/zfChnbMGLVQ?t=494

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=JWGHYUeOx-k

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.

Page

Toolbox

Search

2018 AMC 10B Problems/Problem 23

Contents

Problem

Solution

Video Solution

Video Solution

See Also