2. Jeník wrote several different prime numbers (at least three) on the board. When he added any two of them and then decreased this sum by 7, the resulting number was among those written. Which numbers could have been on the board?
2. Let's denote the prime numbers written on the board as $p_{1}<p_{2}<\ldots<p_{k}$. From these numbers, due to the assumption $k \geqq 3$, it is possible to form $k$ different numbers

$$
p_{1}+p_{2}-7<p_{1}+p_{3}-7<\ldots<p_{1}+p_{k}-7<p_{k-1}+p_{k}-7,
$$

all of which must be among the numbers written on the board. Therefore, they sequentially equal the prime numbers $p_{1}<p_{2}<\ldots<p_{k}$. Precisely, the smallest of them equals $p_{1}$, i.e., $p_{1}+p_{2}-7=p_{1}$, and thus $p_{2}=7$. For the second smallest number in the inequalities (1), we have $p_{1}+p_{3}-7=p_{2}=7$, and thus $p_{1}+p_{3}=14$. For the prime number $p_{1}<p_{2}=7$, we have only three possibilities $p_{1} \in\{2,3,5\}$, which correspond to the values $p_{3}=14-p_{1} \in\{12,11,9\}$, of which only $p_{3}=11$ is a prime number, so $p_{1}=3$. Suppose there is another prime number $p_{4}$ written on the board. Then the third smallest number in the inequalities (1) must be $11=p_{3}=p_{1}+p_{4}-7$, from which, given $p_{1}=3$, it follows that $p_{4}=15$, which is not a prime number.

Answer: Only three prime numbers 3, 7, and 11 could be written on the board.

Let's add that the conclusion $k=3$ can be justified differently by noting that the same $k$-tuple of prime numbers in the group (1) must be obtained even if the last of the numbers is replaced by $p_{2}+p_{k}-7$; thus, it must hold that $p_{k-1}=p_{2}$ or $k=3$.

Another solution. Let's denote the prime numbers written on the board as $p_{1}<p_{2}<\ldots<p_{k}$. If Jeník selects the two smallest and the two largest prime numbers, their sum reduced by 7 is on the board, and thus $p_{1}+p_{2}-7 \geqq p_{1}$ and $p_{k}+p_{k-1}-7 \leqq p_{k}$, from which it follows that $7 \leqq p_{2}<$ $<p_{k-1} \leqq 7$. To avoid a contradiction, we must have $k \leqq 3$. According to the problem, there are at least three prime numbers on the board, so there are exactly three prime numbers $p_{1}<p_{2}<p_{3}$. Since $p_{1}+p_{2}<p_{1}+p_{3}<p_{2}+p_{3}$, it must hold that

$$
p_{1}+p_{2}-7=p_{1}, \quad p_{1}+p_{3}-7=p_{2}, \quad p_{2}+p_{3}-7=p_{3}
$$

From the first (and last) equation, we get $p_{2}=7$ and from the middle equation $p_{1}+p_{3}=14$, with $p_{1}<7=p_{2}$. By testing all possibilities $p_{1} \in\{2,3,5\}$, we get the only solution $p_{1}=3$ and $p_{3}=11$.

For a complete solution, award 6 points, for solving the situation with three prime numbers, award 3 points. Award another 3 points if the student proves that there could be at most 3 prime numbers on the board. For guessing the solution $\{3,7,11\}$, award 2 points.