1996 AIME Problems/Problem 8

Revision as of 16:27, 22 July 2008 by Djkriz (talk | contribs)

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 $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$?

Solution

The harmonic mean of $x$ and $y$ is equal to $2xy/(x+y)$, so we have $xy=(x+y)(3^{20}\cdot2^{19})$, and $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$. $3^{40}\cdot2^{38}$ has $41\cdot39=1599$ factors, one of which is the square root. Since $x<y$, the answer is half of the rest of them, which is $799$.

See also

1996 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS