Decomposing and Recomposing

Let's call a partition of any integer $n$ maximal if the product of the elements of the partition is maximal. (e.g. for $n=4$ we have the three following partitions : $\{1,1,1,1\}$ , $\{1,1,2\}$ , $\{1,3\}$ as well as two maximal partitions : $\{2,2\}$ and $\{4\}$ .)

• It is obvious that $1$ cannot be an element of a maximal partition because we could then get rid of it by adding $1$ to another element, therefore keeping the sum intact and inscreasing the value of the product.

• In the same way, we can discard any partition with an element $p \geq 4$ because then we could split it into $p-2$ and $2$ therefore keeping the sum intact and substituting the corresponding factor in the product value from $p$ to $2(p-2)$ which is always $\geq p$ . We now just proved that there is a maximal partition with only $2$ s and $3$ s.

• We can then show that no more than two $2$ can be elements of the maximal partition : otherwise we could replace $2+2+2$ by $3+3$ therfore keeping the sum intact while increasing a factor in the product from $2\times 2\times 2$ to $3 \times 3$ .

• Since $2018 = 3\times 672 +2$ , we get a maximum value for the product at $2\cdot 3^{672}$ . Hence $a+b+c = 2+3+672 = 677$ .

Since the product of the integers can be expressed as $a\times b^c$ we deduce that $a+bc=2018$

To maximise $a \times b^c$ we introduce the auxiliary function

$\mathcal{L}(a, b, c, \lambda) := ab^c - \lambda (a+bc-2018)$

It follows that

$\displaystyle \frac{\partial \mathcal{L} }{\partial a}= \frac{\partial \mathcal{L} }{\partial b}= \frac{\partial \mathcal{L} }{\partial c}=0$

Solving the equations,

$\begin{aligned}&\frac{\partial \mathcal{L} }{\partial a} = b^c - \lambda = 0 \\ &\frac{\partial \mathcal{L} }{\partial b} = acb^{c-1} - c\lambda =0 \\ &\frac{\partial \mathcal{L} }{\partial c} = ab^c\ln b - b\lambda = 0\end{aligned}$

we have

$b^c = ab^{c-1}=ab^{c-1}\ln b$

Since $a, b, c \in \mathbb{Z^{+}}$ , it is clear that $a\approx b \approx e=2.718\dots$

Which implies that the possible values for $a, b$ should be $2$ or $3$

if $a = 3$ and $b=2$ then $\displaystyle c = \frac{2018-3}{2} = 1007.5 \notin \mathbb{Z}$

if $a = b=3$ then $\displaystyle c = \frac{2018-3}{3} = 671.66\dots \notin \mathbb{Z}$

if $a = 2$ and $b = 3$ then $\displaystyle c = \frac{2018-2}{3} = 672 \in \mathbb{Z}$

if $a=b=2$ then $c = \displaystyle \frac{2018-2}{2} = 1008 \in \mathbb{Z}$

Since $2^{1008} < 3^{672}$ we deduce that $(a, b, c) = (2, 3, 672)$ is the correct answer

Finally $a+b+c=2+3+672=\boxed{677}$

Another way to see that factoring with 3 as much as possible leads to maximal product.

Considering the function $x^{n/x}$ , this has a maximum $x=e$ , the closest integer 3 happens to give the maximum for integers.

So $2018=3*672+2$ leads to $2*3^{672}$

"We now just proved that only $2$ and $3$ can be elements of a maximal partition." - this isn't true. For instance, $\{| 4 |\}$ is a maximal partition of $4$ . (But what can be said is that we can disregard any partition containing a $4$ , since if such a partition is maximal then there is also a maximal partition without any $4$ s in.)