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−2nx−2ny=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