4. Determine the maximum product of natural numbers whose sum is equal to a given natural number $n$.
Solution. Let's denote the maximum product by $P$. For $n=1$, it is obvious that $P=1$. Let's assume that $n>1$. It is clear that none of the factors of the product $P$ is equal to 1. We will prove that all its factors are less than or equal to 4. Indeed, if any of its factors is greater than or equal to 5, then since $2(a-2)>a$, the product which has factors 2 and $a-2$ instead of $a$ will be greater than $P$, and the corresponding sum will be equal to $n$. Furthermore, any fours can be replaced by two twos in both the sum and the product, so we conclude that the product is of the form $2^{a} 3^{b}$. Finally, since $2^{3}<3^{2}$, and $2+2+2=3+3$, we conclude that there can be at most two twos.

From the above, it follows that:

$$
P= \begin{cases}3^{k}, & n=3 k, k \in \mathbb{N} \\ 2^{2} 3^{k-1}, & n=3 k+1, k \in \mathbb{N} \\ 2 \cdot 3^{k}, & n=3 k+2, k \in \mathbb{N} \cup\{0\}\end{cases}
$$

## 4th year