21st Putnam 1960

For n a positive integer find f(n), the number of pairs of positive integers (a, b) such that ab/(a + b) = n.

Solution

ab = n(a + b) implies a(b - n) = nb > 0, so a and b must both exceed n. Let b = n + d, then a = n(n+d)/d = n + n²/d. This is a solution iff d divides n². So f(n) = the number of divisors of n².

21st Putnam 1960

© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002