AMC10 2018 B

Answer (B): Recall that $a\cdot b=\gcd(a,b)\cdot\operatorname{lcm}(a,b)$. Let $x=\operatorname{lcm}(a,b)$ and $y=\gcd(a,b)$. The given equation is then $xy+63=20x+12y$, which can be rewritten as $$(x-12)(y-20)=240-63=177=3\cdot 59=1\cdot 177.$$ Because $x$ and $y$ are integers, one of the following must be true: - $x-12=1$ and $y-20=177$, - $x-12=177$ and $y-20=1$, - $x-12=3$ and $y-20=59$, - $x-12=59$ and $y-20=3$. Therefore $(x,y)$ must be $(13,197)$, $(189,21)$, $(15,79)$, or $(71,23)$. Because $x$ must be a multiple of $y$, only $(x,y)=(189,21)$ is possible. Therefore $\gcd(a,b)=21=7\cdot 3$, and $\operatorname{lcm}(a,b)=189=7\cdot 3^3$. Both $a$ and $b$ are divisible by $7$ but not by $7^2$; one of $a$ and $b$ is divisible by $3$ but not $3^2$, and the other is divisible by $3^3$ but not $3^4$; and neither is divisible by any other prime. Therefore one of them is $7\cdot 3=21$ and the other is $7\cdot 3^3=189$. There are 2 ordered pairs, $(a,b)=(21,189)$ and $(a,b)=(189,21)$.