5. Calculate the maximum product of all natural numbers whose sum is 2019.
Solution. Let us denote the maximum product by $P$. It is obvious that none of the factors in the product $P$ is 1. Since $2(k-2)>k$ and $2+k-2=k$ when $k>4$, it is clear that all factors in the product $P$ are less than or equal to 4 (otherwise, $P$ would not be maximal). Since $4=2+2$, the product does not change when fours are replaced by two twos. It follows that $P=2^{x} 3^{y}$. Since $2^{3}<3^{2}$ and $2+2+2=3+3$, it is clear that there cannot be more than two twos. From $2019=673 \cdot 3$, it follows that $P=3^{673}$.

## IX Section