### IMO 1976

**Problem B1**
Determine the largest number which is the product of positive integers with sum 1976.

**Solution**

Answer: 2·3^{658}.

There cannot be any integers larger than 4 in the maximal product, because for n > 4, we can replace n by 3 and n - 3 to get a larger product. There cannot be any 1s, because there must be an integer r > 1 (otherwise the product would be 1) and r + 1 > 1.r. We can also replace any 4s by two 2s leaving the product unchanged. Finally, there cannot be more than two 2s, because we can replace three 2s by two 3s to get a larger product. Thus the product must consist of 3s, and either zero, one or two 2s. The number of 2s is determined by the remainder on dividing the number 1976 by 3.

1976 = 3·658 + 2, so there must be just one 2, giving the product 2·3^{658}.

Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

18th IMO 1976

© John Scholes

jscholes@kalva.demon.co.uk

10 Oct 1998