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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 $\boxed{\text{ 2}}$ solutions. (awesomeag) Edited by IronicNinja, Firebolt360, and mprincess0229~