9. (1973 Kyiv Mathematical Olympiad) Find three prime numbers such that their product is five times their sum.
9. Let three prime numbers be $p, q, r$, then
$$
p \cdot q \cdot r=5(p+q+r) \text {, }
$$

Thus, one of $p, q, r$ must be 5. Without loss of generality, let $r=5$, then
$$
\begin{array}{l}
5 p q=5(p+q+5), \\
p q=p+q+5, \\
(p-1)(q-1)=6 .
\end{array}
$$

Since $p>1, q>1$, we have
$$
\left\{\begin{array} { l } 
{ p - 1 = 2 , } \\
{ q - 1 = 3 , }
\end{array} \left\{\begin{array} { l } 
{ p - 1 = 3 , } \\
{ q - 1 = 2 , }
\end{array} \left\{\begin{array} { l } 
{ p - 1 = 1 , } \\
{ q - 1 = 6 , }
\end{array} \left\{\begin{array}{l}
p-1=6, \\
q-1=1 .
\end{array}\right.\right.\right.\right.
$$

Since $q=4$ or $p=4$ is not a prime number, the only solutions are $p=2, q=7$, or $p=7, q=2$. Therefore, the three prime numbers are 2, 5, 7.