1. On the board, there are written (not necessarily distinct) prime numbers, whose product is $105k$ times greater than their sum. Determine all the prime numbers written on the board if there are

(a) five;

(b) seven.

(Tomáš Jurík, Jaromír Šimša)
SOLUTION. (a) For the prime numbers $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}$ written on the board according to the problem statement, we have

$$
p_{1} p_{2} p_{3} p_{4} p_{5}=105\left(p_{1}+p_{2}+p_{3}+p_{4}+p_{5}\right).
$$

Since the number on the right side (1) is a multiple of the number $105=3 \cdot 5 \cdot 7$, three of the five prime numbers on the left side (1) must be 3, 5, and 7. Without loss of generality, we can assume that $p_{3}=3, p_{4}=5$, and $p_{5}=7$. Equation (1) then simplifies to

$$
p_{1} p_{2}=p_{1}+p_{2}+15 \text {, }
$$

after substitution and division of both sides by 105. Such an equation with unknown integers $p_{1}, p_{2}$ can be transformed into a product form by standard manipulation:

$$
\left(p_{1}-1\right)\left(p_{2}-1\right)=16 \text {. }
$$

In our problem, $p_{1}, p_{2}$ are prime numbers, so both factors $p_{1}-1$ and $p_{2}-1$ are natural numbers. We can certainly choose the labeling such that $p_{1} \geqq p_{2}$, and thus $p_{1}-1 \geqq p_{2}-1$.

The number 16 on the right side (2) can be factored into two natural numbers $a=p_{1}-1$, $b=p_{2}-1$ satisfying the condition $a \geqq b$ in exactly these ways: $16 \cdot 1, 8 \cdot 2$, and $4 \cdot 4$. These factorizations correspond to pairs $\left(p_{1}, p_{2}\right)$ equal to $(17,2),(9,3),(5,5)$. Only the middle pair is not a pair of prime numbers. By adding the two extreme pairs to the triplet $(3,5,7)$ from the initial part of the solution, we get the only two solutions to part (a) of the problem, which we now write for clarity as a quintuple of prime numbers in non-decreasing order: $2,3,5,7,17$ and $3,5,5,5,7$.

(b) In this case, for the prime numbers $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ written on the board, the equation

$$
p_{1} p_{2} p_{3} p_{4} p_{5} p_{6} p_{7}=105\left(p_{1}+p_{2}+p_{3}+p_{4}+p_{5}+p_{6}+p_{7}\right)
$$

must hold. Similarly to part (a), we conclude that without loss of generality, $p_{5}$, $p_{6}$, and $p_{7}$ are the prime numbers 3, 5, and 7, and we are left to solve the equation

$$
p_{1} p_{2} p_{3} p_{4}=p_{1}+p_{2}+p_{3}+p_{4}+15.
$$

We can certainly assume that $p_{1}$ is the largest (or one of the largest) among the numbers $p_{1}, p_{2}, p_{3}, p_{4}$. Since they are prime numbers, each of them is at least 2. Thus, we can estimate the left and right sides of equation (3) as follows:

$$
p_{1} \cdot 2 \cdot 2 \cdot 2 \leqq p_{1} p_{2} p_{3} p_{4}=p_{1}+p_{2}+p_{3}+p_{4}+15 \leqq 4 p_{1}+15
$$

For the modified extreme terms, we have the inequality $8 p_{1} \leqq 4 p_{1}+15$, from which $p_{1} \leqq 3.75$, so for the prime number $p_{1}$, we have $p_{1} \in\{2,3\}$. Since we chose $p_{