Suppose that we have some positive integers (not necessarily distinct) whose sum is \(100\). How large can their product be? You should prove your answer is best.

I got solutions from **Alice Butler**, **Mohammad Tayyab Sajjad**, **Ben Andrews**, **James Mason** and **Oliver Feghali**. Mostly they were slightly incomplete as proofs, but they had all of the ingredients: by combining bits and pieces I can produce a really nice one.

The best solution can't use a 1, as \(1+n\) is always bigger than \(1\times n\).

The best solution also can't use anything larger than 4, as \(2(n-2)\) is bigger than \(2+(n-2)\) for \(n\geq 5\); similarly, it needn't use a 4, as \(2\times 2\) is equal to \(2+2\).

Therefore, we can only use numbers 2 and 3.

There can be at most two 2's, since \(3\times 3 > 2\times 2\times 2\).

So we make our number with 3's, and either zero, one or two 2's. The only way of making a hundred is with thirty-two 3's and two 2's. This gives an answer of \(2^23^{32}\).