2. On the board, there are several (not necessarily distinct) prime numbers such that their product is $2020$ times greater than their sum. Determine their smallest possible number. (Patrik Bak)
SOLUTION. Let the prime numbers on the board be denoted as $p_{1}, p_{2}, \ldots, p_{n}$. According to the problem statement, we have

$$
2020 \cdot\left(p_{1}+p_{2}+\ldots+p_{n}\right)=p_{1} p_{2} \ldots p_{n}
$$

The left side of equation (1) is divisible by the number $2020=2 \cdot 2 \cdot 5 \cdot 101$. Therefore, the prime numbers $2,2,5,101$ must appear in the prime factorization on the right side of (1). Without loss of generality, assume that $p_{1}=2, p_{2}=2, p_{3}=5, p_{4}=101$. If only these 4 prime numbers were on the board (case $n=4$), then equation (1) would clearly not hold $(2020 \cdot 110 \neq 2020)$. Therefore, it must be that $n \geqq 5$.

By substituting the already known prime numbers $p_{1}=2, p_{2}=2, p_{3}=5, p_{4}=101$ into equation (1) and dividing by 2020, we get the equation

$$
110+p_{5}+p_{6}+\ldots+p_{n}=p_{5} \cdot p_{6} \cdot \ldots \cdot p_{n}
$$

Since we are looking for the smallest possible $n \geqq 5$ for which there exist prime numbers $p_{5}, p_{6}, \ldots, p_{n}$ satisfying equation (2), we will further analyze the smallest possible values of $n=5,6, \ldots$, until we find the first suitable $n$.

- Case $n=5$. Equation (2) is of the form $110+p_{5}=p_{5}$, which is not satisfied by any prime number $p_{5}$.
- Case $n=6$. The corresponding equation is

$$
p_{5} p_{6}=p_{5}+p_{6}+110
$$

which we rearrange into the product form

$$
\left(p_{5}-1\right)\left(p_{6}-1\right)=111
$$

Both factors on the left side must be odd, so the numbers $p_{5}$ and $p_{6}$ must both be even.* The only even prime numbers $p_{5}=p_{6}=2$ do not satisfy equation (3).

- Case $n=7$. Equation (2) has the form

$$
p_{5} p_{6} p_{7}=p_{5}+p_{6}+p_{7}+110 .
$$

Let us emphasize that our task is only to determine whether there exists a triplet of prime numbers that satisfies equation (4). It is not difficult to guess the triplet $p_{5}=p_{6}=p_{7}=5$:

$$
5^{3}=125=5+5+5+110 .
$$

This completes the solution of the problem. Let us add that other triplets of prime numbers that satisfy equation (4) are $(2,5,13)$ and $(2,3,23)$.** Therefore, if we do not guess the triplet $(5,5,5)$, we can complete the solution by substituting the smallest existing prime number $p_{5}=2$ into equation (4). For the prime numbers $p_{6}$ and $p_{7}$, we then get the equation, which, when converted to the product form $\left(2 p_{6}-1\right)\left(2 p_{7}-1\right)=225$, yields two prime number solutions $(3,23)$ and $(5,13)$.

Summarizing our findings: At least 7 prime numbers were written on the board, an example of such a septet (arranged in ascending order) is $2,2,5,5,5,5,101$.

Conclusion. The smallest possible number of prime numbers on the board is 7. [^0]NOTES.