AIME 1996 · 第 8 题

Solution

The harmonic mean of $x$ and $y$ is equal to $\frac{1}{\frac{\frac{1}{x}+\frac{1}{y}}2} = \frac{2xy}{x+y}$, so we have $xy=(x+y)(3^{20}\cdot2^{19})$, and by SFFT, $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$. Now, $3^{40}\cdot2^{38}$ has $41\cdot39=1599$ factors, one of which is the square root ($3^{20}2^{19}$). Since $x, the answer is half of the remaining number of factors, which is$\frac{1599-1}{2}= \boxed{799}$.

Note: Counting how many positive integer divisors of $n^2$ less than $n$ is also used in 1995 AIME Problems/Problem 6.