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Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes
 
Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes
  

Revision as of 16:58, 16 February 2018

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

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

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 $(189, 21)$. Obviously, the first pair does not work. The second pair can be ordered in two ways. Thus, the answer is $\boxed{2}$. (awesomeag)

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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