4.1. The product of several (not necessarily distinct) prime numbers is 10 times greater than their sum. Find these numbers.
4.1. It is clear that among our numbers there are 2 and 5. Let the others be $p_{1}, p_{2}, \ldots, p_{k}$. Then denote $P=p_{1} p_{2} \ldots p_{k}=$ $7+p_{1}+p_{2}+\ldots+p_{k}$. If $k \geqslant 4$, then $p_{i} \leqslant P / 8(i=1,2, \ldots, k)$ and $7 \leqslant 7 P / 16<P / 2$. Adding these inequalities, we get that the right side is less than $P$ - a contradiction. For $k=1$, we get $7=0$, which is not true. For $k=2$, we have $p_{1} p_{2}=7+p_{1}+p_{2}$, $\left(p_{1}-1\right)\left(p_{2}-1\right)=8, p_{1}=3, p_{2}=5$ (or vice versa). For $k=3$, we get $p_{1} p_{2} p_{3}=7+p_{1}+p_{2}+p_{3}$. If all $p_{i}$ are odd, then the left and right sides have different parity. If, say, $p_{3}=2$, then by parity one of the numbers $p_{2}, p_{1}$ is also 2, and for the remaining we get the equation $4 p=p+11$, which has no solutions in integers. Thus, the answer: $\{2,3,5,5\}$.