5. A natural number $n$ is such that the sum of four of its distinct natural divisors (possibly including the number itself) is equal to $2 n$. What can the sum of the four smallest natural divisors of this number be? List all possible answers and prove that there are no others.
Solution. The sums can be 10, 11, and 12.

We will prove that the divisors that sum up to $2n$ are $n, n/2, n/3$, and $n/6$.

Indeed, if there is no divisor $n$, then the largest possible sum is

$$
\frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}<2n
$$

If there is no divisor $n/2$, then the maximum sum is

$$
n+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}<2n
$$

If there is no divisor $n/3$, then the maximum sum is

$$
n+\frac{n}{2}+\frac{n}{4}+\frac{n}{5}<2n.
$$

The remaining divisor is

$$
2n-n-\frac{n}{2}-\frac{n}{3}=\frac{n}{6} \text{.}
$$

Thus, the number $n$ has divisors $1, 2, 3, 6$. There may also be (or may not be) divisors 4 and 5. Therefore, the sum of the smallest divisors can be $1+2+3+4=10$ (for example, for $n=12$), $1+2+3+5=11$ (for $n=30$), and $1+2+3+6=12$ (for $n=6$).