[ Number of divisors and sum of divisors of a number ] equations in integers $\quad]$

Find a natural number $n$, knowing that it has two prime divisors and satisfies the conditions $\tau(n)=6$, $\sigma(n)=28$.
According to the formulas from problem $\underline{60537} n=p^{2} q,\left(p^{2}+p+1\right)(q+1)=28=22 \cdot 7$, where $p$ and $q-$ are prime numbers. From this, $q$ $=3, p=2$.

## Answer

$n=12$.