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 $x<y$ 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 $x\geq y$ , the answer is half of the rest of them, which is $799$ .

1996 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1996 AIME Problems/Problem 8

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