Split 2018

Note that $2(a-2) > a$ when $a > 4$ , and that $a+1 > a\times1$ . Also note that $2^3 < 3^2$ , while $2+2+2=3+3$ .

Firstly, there are only finitely many ways of writing a positive integer $N$ as a sum of positive integers (there can only be at most $N$ such integers, and none of them can exceed $N$ ). Thus there does indeed exist a maximum integer $M$ such that $M$ can be written as a product of positive integers which add to $N$ . Suppose that $a_1,a_2,a_3,...,a_n$ are positive integers that add to $N$ and multiply to $M$ .

If $a_1 > 4$ then we could replace $a_1$ by $a_1-2$ and $2$ , obtaining another set of positive integers that add to $N$ , but with a larger product. Since this is not possible, we deduce that $a_1 \le 4$ . Similarly, $a_j \le 4$ for all $j$ . If $a_1 = 1$ we could discard $a_1$ and replace $a_2$ by $a_2+1$ , obtaining a larger product. This is not possible, and so $a_1 \ge 2$ . Similarly $a_j \ge 2$ for all $j$ . Since $4 = 2+2 = 2\times2$ , we can, without loss of generality, assume that no $a_j$ is equal to $4$ (otherwise we could replace it with two $2$ s). Thus we deduce that $a_j = 2,3$ for all $j$ . Finally, since $2^3 < 3^2$ , we deduce that there can only be at most two $2$ s in the list. To sum up, the list of numbers $a_1,a_2,...,a_n$ can only contain $2$ s and $3$ s, and can contain at most two $2$ s in addition.

If $N = 3m$ , then the only way these conditions can be satisfied is if $n = m$ and $a_j = 3$ for all $1 \le j \le m$ , making $M = 3^m$ .

If $N = 3m+1$ , then the only way these conditions can be satisfied is if $n = m+1$ and $a_j = 3$ for $1 \le j \le m-1$ , with $a_m=a_{m+1} = 2$ (or a reordering of these numbers), making $M = 4 \times 3^{m-1}$ . In this case, it is also possible that $n=m$ , with $a_j=3$ for $1 \le j \le m-1$ and $a_m = 4$ , up to reordering.

If $N = 3m+2$ , then the only way these conditions can be satisfied is if $n=m+1$ and $a_j=3$ for $1 \le j \le m$ , with $a_{m+1} = 2$ (or a reordering of these numbers), making $M = 2 \times 3^m$ .

In the case $N = 2018 = 3\times672+2$ , we obtain $n = 672+1=\boxed{673}$ .

@Mark Hennings ' explanation is way better, but the way I did it was by treating $n$ and all the terms as any positive real number instead of just integers and finding that, firstly, for any given $n$ the product is maximized when all terms are the same value, which means the product is $p=\left(\frac{N}{n}\right)^n$ , and secondly that this product is maximized when $n=\frac{N}{e}$ --i.e. each term is $e$ , which is between 2 and 3; closer to 3. This is enough to make an educated guess by making as many terms as possible equal 3 and letting the other two be 2, so since we had 3 tries, that is basically what I did. See comments below for details.

For any set of positive numbers that have a fixed sum, the maximum product is achieved when they are the same number (or as close to being the same as possible). An excellent/ probably better proof for this is given by @Eric Nordstrom but the way I think of it is (perhaps more intuitive/visual) given the perimeter of the sides of a shape, the maximum volume/area is achieved when the sides are the same length.

An example for n=2 : Maximise the area of the rectangle with side lengths a and b, given that the perimeter is 2x.

We know a + b = x and we need to maximise ab. Through substitution: ab = a(x-a) , which has a maximum at a= x/2 so b = x/2. This can also be shown for larger n.

Therefore a good start is to let all the numbers equal $\frac{2018}{n}$ so we need to maximise ( $\frac{2018^{n}}{n^{n}}$ )

$\frac{d}{dn}$ ( $\frac{2018^{n}}{n^{n}}$ ) = $\frac{ln(2018) \times 2018^{n} - 2018^{n} \times (ln(n)+1)}{n^{n}}$ solving for $\frac{d}{dn}$ ( $\frac{2018^{n}}{n^{n}}$ ) = 0, gives n = $\frac{2018}{e}$ so the maximum value is given when n = 742.3807123... and all the numbers in the set are e but n and a must be integers so we need to let as many integers be as close to e as possible.

Since 3 is the closest integer to e, we let as many numbers be 3 before the sum exceeds 2018: so there are 672 3's in the set. $672 \times 3$ = 2016 so we can have 2 1's or a 2 but 2> $1 \times 1$ so we choose 2 and hence there are 673 numbers in the set