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

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<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath>
 
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath>
  
−
Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as
+
Using [[Simon's Favorite Factoring Trick]]or SFFT, we rewrite this equation as
  
 
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath>
 
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath>

Revision as of 15:35, 27 January 2020

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 $x = lcm(a, b)$, 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 Trickor SFFT, 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 $\text{GCD}$ 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 $\boxed{2}$ solutions. (awesomeag)

Edited by IronicNinja, Firebolt360, and mprincess0229~

Video Solution

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

See Also

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

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