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Maximum product of integers that sum to a fixed number
By Philip Ellison on Wednesday, December 11, 2002 - 07:01 am:

An integer n is divided into integer parts.(e.g. 10 = 6 + 3 + 1 so the parts are 6, 3 and 1.) What is the best way of dividing the number so that the product of the parts will be as large as possible? I believe that for numbers of the form 3k, the maxiproduct is 3n, for numbers of the form 3k+1, the maxiproduct is 3(n-4)/3.22 and for numbers of the form 3k+2, the maxiproduct is 3(n-2)/3.2. My reasoning is that as 32>23, 35>53, etc. it is always preferable to use 3 rather than any other prime. Is there a nicer way of showing this (and is my reasoning correct?).
Thanks

By Saul Foresta on Wednesday, December 11, 2002 - 07:39 am:

Yea that sounds about right. I would reason as follows, since, we wish to have a maximal product, none of the integers should be greater than 4, because if this were so then we could replace that integer, call it r by r-2 and 2, whose product is 2r-4 which is greater than r for r>4. Also there should be no more than 2 2s because we may replace 3 2s by 2 3s to acheive a greater product. Finally, as you have done i would just break the numbers n into 3 cases, those which are 0 mod 3, 1 mod 3, and those which are 2 mod 3.

By Demetres Christofides on Wednesday, December 11, 2002 - 09:41 am:

Both reasonings were correct. I would add the following to Saul's reasoning.

1) It is not necessary to have any 4's since we can replace each 4 be 2 2's (leaving both the sum and the product unchanged) [If the number is of the form 3k+1 (k>0) you can also have 3a.4 instead of 223a]
2) Trivial but still need to say it: None of the integers should be 1 otherwise replace 1 and r by r+1>1.r [Of course this only works if the sum is >1]

I think this appeared in some of the early IMO's (when they used to be much easier than today)

Demetres