7-0. The number $n$ has exactly six divisors (including 1 and itself). They were arranged in ascending order. It turned out that the third divisor is seven times greater than the second, and the fourth is 10 more than the third. What is $n$?
Answer: 2891

Solution. If $n$ has six divisors, then either $n=p^{5}$ or $n=p \cdot q^{2}$ (where $p$ and $q$ are prime numbers). In the first case, $p=7$ (since the third divisor is seven times the second), but then the second condition is not satisfied.

Therefore, $n$ has two prime divisors (one of which is squared), and one of the prime factors is 7. Let the smallest divisors of $n$ be $1, p, 7p$, and $7p+10$. If $p=7$, then $7p+10=59$, and we get the answer 2891. If $p<7$, then $p=2$ (then $7p+10=24$, and such a number has more than six divisors). If $p=3$, then $7p+10=31$, and if $p=5$, then $7p+10=45$. In each of these cases, $n$ has more than two prime divisors.