## Task 26/70

We are looking for all natural numbers $n$ that have exactly 10 divisors.
The sought numbers $n$ can contain at most two different prime factors $p$ and $q$. It is easy to verify that with three different prime factors $p, q$, and $r$, eight (if all appear in the first power) or more than 10 divisors result; with more than four prime factors, the number of divisors is always greater than 10.

If $n$ contains exactly one prime factor $p$, this prime factor must be present in the 9th power: $n=p^{9}$. The 10 divisors are then the 10 powers $p^{0}, p^{1}, p^{2}, \ldots, p^{9}$.

If $n$ contains exactly two different prime factors $p$ and $q$, one of them can only appear in the first power. If it were in the $k$-th power with $k>1$, it would already yield $k+1$ divisors.

Since each of these divisors, when multiplied by a divisor resulting from the other prime factor, again yields a divisor of $n$, the number of divisors of $n$ must be a multiple of $k+1$. Therefore, in our case, only 1 and 4 can be the exponents of the prime factors $p$ and $q$. Indeed, $p^{4} q$ has exactly 10 divisors.

The sought numbers $n$ thus have the prime factorizations $n=p^{9}$ and $n=p^{4} q$ with $p \neq q$.