## Task B-3.6.

A natural number $n$ has a total of 6 divisors, of which exactly two are prime numbers. Determine the number $n$ if the sum of all its divisors is equal to 248.
## Solution.

(If the number $n$ is of the form $n=p \cdot q$ where $p$ and $q$ are prime numbers, then the number $n$ has exactly 4 divisors. These are the numbers $1, p, q, p q$.

The number $n$ is of the form $p^{2} \cdot q$ and its divisors are $1, p, p^{2}, q, p q, p^{2} q$.

Since the sum of all divisors of the number $n$ is equal to 248, we have

$$
1+p+p^{2}+q+p q+p^{2} q=248
$$

It follows that

$$
\begin{gathered}
q\left(1+p+p^{2}\right)+\left(1+p+p^{2}\right)=248 \\
\left(1+p+p^{2}\right)(q+1)=248 \\
\left(1+p+p^{2}\right)(q+1)=2^{3} \cdot 31
\end{gathered}
$$

Since the number $1+p+p^{2}=1+p(p+1)$ is an odd number, it must be $1+p+p^{2}=31$ and thus $q+1=8$.

Therefore, $q=7$, and from $p^{2}+p+1=31$ we get $p=5$ (the negative solution $p=-6$ is discarded).

Thus, $n=5^{2} \cdot 7=175$.