Solution (Prof. Cristian Mangra, Bucharest):

Answer: $n \in \mathbb{N}^{*}-\{1,2,3,5\}$

- If $n$ is a composite number, $n=p q, p>1, q>1$, we take $a_{1}=p, a_{2}=q$ and $a_{3}=a_{4} \ldots=a_{k}=1$, where $k=n-(p+q)$. Thus, it is verified that $a_{1}+a_{2}+\ldots+a_{k}=a_{1} a_{2} \cdot \ldots \cdot a_{k}=n$.
- If $n$ is a prime number $n \geq 11$, we take $a_{1}=\frac{n}{2}, a_{2}=\frac{1}{2}, a_{3}=4$ and $a_{4}=a_{5}=\ldots=a_{k}=1$, where $k=\frac{n-3}{2}$. Thus, it is verified that $a_{1}+a_{2}+\ldots+a_{k}=a_{1} a_{2} \cdot \ldots \cdot a_{k}=n$.
- If $n=7$, we take $k=3$ and $a_{1}=\frac{9}{2}, a_{2}=\frac{7}{6}, a_{3}=\frac{4}{3}$.

Assume there exists a natural number $n$, less than or equal to 5, which is not composite and satisfies the statement.

From the inequality of means, we get $\frac{a_{1}+a_{2}+\ldots+a_{k}}{k} \geq \sqrt[k]{a_{1} a_{2} \ldots a_{k}}$, equivalent to $n^{k-1} \geq k^{k}$.

- The cases $n \in\{1,2\}$ do not work.
- If $n=3$, we have $3^{k-1}<k^{k}$, for any $k \geq 2$. It does not satisfy $n^{k-1} \geq k^{k}$.
- If $n=5$, we have $5^{k-1}<k^{k}$, for any $k \geq 3$. It does not satisfy $n^{k-1} \geq k^{k}$.

For $k=2$, we get $a_{1}+a_{2}=a_{1} a=5$, which does not hold for rational numbers.
## Problem 3

For the non-zero natural number $n$, there exists a natural number $k, k \geq 2$, and positive rational numbers $a_{1}, a_{2}, \ldots, a_{k}$ such that $a_{1}+a_{2}+\ldots+a_{k}=a_{1} a_{2} \cdot \ldots \cdot a_{k}=n$. Determine all possible values of the number $n$.