### 40th Putnam 1979

**Problem A1**

Find the set of positive integers with sum 1979 and maximum possible product.

**Solution**

*Easy*

For n > 4, 2(n - 2) > n, so the maximum product cannot include any integer greater than 4. Also, 2^{3} < 3^{2}, so it cannot include more than two 2s. Since 4 = 2·2, it cannot include both a 2 and a 4. It obviously does not include a 1, since n + 1 > n x 1. So the maximum product must be made up mainly of 3s, with either no, one or two 2s (or equivalently one 4). Hence for 1979 = 2 + 659·3, the maximum product is 3^{659}2.

40th Putnam 1979

© John Scholes

jscholes@kalva.demon.co.uk

3 Nov 1999