The harmonic mean of two positive numbers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x, y)$ with $x < y$ is the harmonic mean equal to $6^{20}$?
I don't really know how to go about this problem..
Let $n=\frac{6^{20}}{2} $, note that the harmonic mean is $\frac{2 x y}{x+y}$, then the equation can be written as : $$n(x+y)=xy\Longleftrightarrow (x-n)(y-n)=n^2 .$$ if the number of divisors of $n^2$ is $s$ then the answer is $\lfloor \frac{s}{2}\rfloor$ (since $x>y$).
In this case $n=2^{19}\cdot 3^{20}$, then $n^2=2^{38}3^{40}$, and the $s=39\cdot41$, the answer is then $\boxed{758}$.