Difference between revisions of "2018 AMC 10B Problems/Problem 23"

m (Solution)
Line 21: Line 21:
  
 
Edited by IronicNinja~
 
Edited by IronicNinja~
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=JWGHYUeOx-k
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:29, 11 February 2019

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

\[(x - 12)(y - 20) - 240 + 63 = 0\] \[(x - 12)(y - 20) = 177\]

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 $a<b$, so there are $\boxed{2}$ solutions. (awesomeag)

Edited by IronicNinja~

Video Solution

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

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS