Problem:
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 620?
Solution:
Let n=620. Suppose that x and y are positive integers for which
n=21​(x1​+y1​)1​=x+y2xy​
It follows that xy−2n​x−2n​y=0, hence that
(x−2n​)(y−2n​)=4n2​=4640​=238340
Because 238340 has 39⋅41=1599 positive divisors, there are 21598​=799 pairs of unequal positive integers whose product is 238340, and therefore 799​ ordered pairs (x,y) of the required type.
The problems on this page are the property of the MAA's American Mathematics Competitions