### 21st Putnam 1960

**Problem A1**

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 + n^{2}/d. This is a solution iff d divides n^{2}. So f(n) = the number of divisors of n^{2}.

21st Putnam 1960

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002