# Difference between revisions of "1996 AIME Problems/Problem 8"

## Problem

The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x is the harmonic mean of $x$ and $y$ equal to $6^{20}$?

## 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 $\frac{1599-1}{2}= \boxed{799}$.

## See also

 1996 AIME (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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