Problem source

Find the integers $k$ satisfying the inequality $k\leqslant2(k-2).$
Given that $N$ is a strictly positive integer consider the problem of finding strictly positive integers whose sum is $N$ and whose product is as large as possible. Call this largest possible product $P(N).$ 
Show that $P(5)=2\times3, P(6)=3^{2}, P(7)=2^{2}\times3, P(8)=2\times3^{2}$ and $P(9)=3^{3}.$
Find $P(1000)$ explaining your reasoning carefully.

Solution source

\begin{align*}
&& k &\leq 2(k-2) \\
\Rightarrow && 4 &\leq k
\end{align*}

Lemma: Suppose we construct $N \neq $ (optimally) as a sum out of $a_1 + \cdots +a_k$, then $a_i \in \{2, 3\}$.

Proof: Suppose not, suppose some $a_i > 3$. Then from our earlier inequality, the sum $a_1 + \cdots +a_{i-1} + 2 + (a_i - 2) + \cdots $ has the same sum, but a larger product. Therefore $a_i \leq 3$. 

Suppose also some $a_i = 1$, then we could replace $a_1$ with $a_1+1$ and remove $a_i$, leaving us again with the same sum but larger product. (Assuming $N \neq 1$)

$5 = 2+3$ is the only way to write $5$ as a sum of $2$s and $3$s, therefore $P(5) = 2\times 3$

$6 = 2 + 2 + 2 = 3 + 3$ and we can immediately see that $2^3 = 8 < 3^2 = 9$, so $P(6) = 3^2$ and whenever we have three $2$s we should replace them with two $3$s. So

$7 = 2 + 2 + 3 \Rightarrow P(7) = 2^2 \times 3$
$8 = 3 + 3 + 2 \Rightarrow P(8) = 2\times 3^2$
$9 = 3 + 3 + 3 \Rightarrow P(9) = 3^3$

Suppose $1000 = 2n + 3m$, considered $\pmod{3}$ we can see that $n \equiv 2 \pmod{3}$ therefore we should have $1000 = 2 + 2 + \underbrace{3 + \cdots + 3}_{332\text{ }3\text{s}}$ and so $P(1000) = 2^2 \times 3^{332}$