Split 2018 Ⅲ

We make some obvious observations: (a) In the solution, there will be no summands $m \geq 7$ since "splitting" $m=1+3+(m-4)$ will increase the product; (b) There may be some 1's or 5's but not both since $1+5=3+3$ ; (c) There will be at most one 5 since $5+5=3+3+3+1$ ; (d) There will be at most two 1's since $1+1+1=3$ .

In the case $N=2018$ , this leaves us with two options, $2018=671\times 3 + 5= 672\times 3 + 2\times 1$ , with the first being the "winner," 5:3. The answer is $\boxed{672}$

Bonus: If $N$ is divisible by 3 (as will happen in a few days' time , in 2019), we use all 3's; if the remainder is 1 we use one 1 and the rest 3's; and if the remainder is 2 (as in 2018) we use one 5 and the rest 3's except for $N=2$ , where we use two 1's.

Note that $1 \times 5 < 3^2$ , $5^2 < 3^3\times1$ , $1^3 < 3^1$ , $3 \times 1^2 < 5$ and also that $n < 9(n-6)$ for all $n \ge 7$ .

If $N$ is written as a sum of positive odd integers so as to maximise their product $M$ , none of the positive integers involved can be $7$ or more (otherwise we could replace $a \ge 7$ in the decomposition by $a-6,3,3$ ). Thus the only odd numbers used are $1,3,5$ . We also note that at most two of the numbers can be $1$ , and at most one of the numbers can be $5$ , and also that we cannot use both $1$ and $5$ at the same time. Moreover, provided that we use at least one $3$ , we cannot use more than one $1$ . Suppose now that $N \ge 6$ , so that we have to use at least one $3$ .

• If $N = 3m$ then we must have no $1$ s and no $5$ s, and hence $M = 3^m$ , making $n = m$ .
• If $N=3m+1$ , then we must have one $1$ and no $5$ s, and hence $M = 3^m \times 1$ , making $n = m+1$ .
• If $N = 3m+2$ , then we must have no $1$ s and one $5$ , and hence $M = 3^{m-1} \times 5$ , making $n = m$ .

In the case $N = 2018$ , we deduce that $n = \boxed{672}$ .

This is just an intuition. knowing the answer is $672$ , for the case where $a_t \ \forall t$ can be any positive integer, and the fact that $a_t \approx 3 \ \forall t$ for the mentioned case, it is easy to guess

$3 \times 672 + 2 = 2018$

Now we need to get rid of the $2$ in the summation. it can be added to one of the $3$ s. so, there would be still $672$ terms to maximise the product.