1996 AIME Problem 8

Question

The harmonic mean of two positive numbers 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}?

Answer

Answer: 799

Concepts:harmonic mean factor counting prime factorization

Difficulty rating: 2380

Small hint:

For

N=6^{20},

rearrange

\frac{2xy}{x+y}=N

into a product

Big hint:

Count complementary factor pairs of

N^2

in which both factors are even

Solution:

Let

N=6^{20}.

Rearranging the harmonic-mean equation gives

(2x-N)(2y-N)=N^2.

Because

x<N<y

and

N<2x,

the two factors are positive, and both must be even. Conversely, each factorization

AB=N^2

with even

A<B

gives one valid pair via

x=\frac{A+N}{2}

and

y=\frac{B+N}{2}.

Now

N^2=2^{40}3^{40}.

For both complementary factors to be even, the exponent of

2

in

A

can be

1,\ldots,39,

while the exponent of

3

can be

0,\ldots,40.

This gives

39\cdot41=1599

divisors

A,

including the central factor

A=N.

Pairing complementary divisors and excluding that central case gives

\frac{1599-1}{2}=799.