1996 AIME Problems/Problem 8

Revision as of 16:26, 22 July 2008 by Djkriz (talk | contribs) (Undo revision 27128 by Djkriz (Talk))

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 $39\cdot41=1599$ 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