An integer has two prime factors. It has 6 divisors, and the sum of its divisors is 28. Which number is this?
Solution. According to the conditions, the number is of the form $p_{1}^{n} \cdot p_{2}^{m}$, where $n$ and $m$ are positive integers and $p_{1} \neq p_{2}$ are primes. Based on the formula for the number of divisors, $(n+1)(m+1)=6$. The pair of numbers $n=2, m=1$ is the only one that satisfies the given conditions, so the number is of the form $p_{1}^{2} \cdot p_{2}$. The divisors of the number are then: $1, p_{1}, p_{2}, p_{1}^{2}, p_{1} \cdot p_{2}, p_{1}^{2} \cdot p_{2}$, that is:

$$
\begin{aligned}
& 28=1+p_{1}+p_{2}+p_{1}^{2}+p_{1} p_{2}+p_{1}^{2} p_{2} \\
& 28-\left(1+p_{1}+p_{1}^{2}\right)=p_{2}\left(1+p_{1}+p_{1}^{2}\right) \\
& p_{2}=\frac{28-\left(1+p_{1}+p_{1}^{2}\right)}{1+p_{1}+p_{1}^{2}}=\frac{28}{1+p_{1}+p_{1}^{2}}-1
\end{aligned}
$$

Since for $p_{1} \geq 5$ the denominator is greater than 28, only $p_{1}=2$ or $p_{1}=3$ are possible. In the first case, $p_{2}=3$, and in the second case, we do not get an integer for $p_{2}$.

Therefore, the sought number is: $2^{2} \cdot 3$, that is, 12.