# Product of Integers That Add Up to 2018

### Problem

### Solution

For an integer $a\ge 4,$ $(a-2)+2=a$ and $(a-2)2\ge a.$ So, if there is a product of positive integers that add up to $2018,$ it is possible to replace the numbers in the product without changing the sum such that each number involved would be $2$ or $3$ while the product would increase.

Now from, $2+2+2=3+3$ and $2^3 = 8\lt 9=3^2,$ we may only increase the product by collecting triples of $2s$ and replacing them with pairs of $3s.$

$2018=3\cdot 672+2.$ It appears that the maximum product equals $3^{672}\cdot 2.$

### Acknowledgment

This is a slight modification of a problem from an olympiad chapter of *Arbelos* (1983, v 2, n 1, p 19-20), a now defunct magazine published by Samuel L. Greitzer.

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny64669322 |