## Task 11/83

We are looking for all natural numbers $n$ for which the sum $s$ of the proper divisors is equal to 39.
First, we bound the number of distinct prime factors of $n$ from above. Suppose $n$ contains three distinct prime factors: $n=p_{1}^{k} p_{2}^{l} p_{3}^{m}$. Even the smallest such number $(n=2 \cdot 3 \cdot 5=30)$
already contains the divisors $2, 3, 5, 6, 10,$ and $15$, and thus has the divisor sum $s=41>39$.

Therefore, $n$ can have at most two distinct prime factors $p_{1}$ and $p_{2}$. Thus, only $n=p_{1}^{k} p_{2}^{l}$ with $k \geq 0$ is possible.

Let's further bound $k$ and $l$. For $k=l=2$, the smallest primes $p_{1}=2, p_{2}=3$ already yield the divisor sum $s=51>39$; thus, at least one of the two exponents must be less than 2. If $k=l=1$, then $s=p_{1}+p_{2}=39$ precisely when one of the prime factors is 2 and the other is 37. Thus, one solution is found: $n=2 \cdot 37=74$.

Now, we need to investigate the cases

$$
n=p_{1}^{k} p_{2} \quad(1) \quad \text { and } \quad n=p_{1}^{k}
$$

(i.e., $l=1, k>1$ and $l=0, k>1$).

For (1): We have

$$
s=p_{1}+p_{1}^{2}+\ldots+p_{1}^{k}+p_{2}+p_{1} p_{2}+p_{1}^{2} p_{2}+\ldots+p_{1}^{k-1} p_{2}=\left(p_{1}+p_{1}^{2}+\ldots+p_{1}^{k-1}\right)\left(1+p_{2}\right)+p_{2}+p_{1}^{k}>p_{1}^{k}
$$

Consequently, due to $p_{2} \geq 2$, only prime powers $p_{1}^{k}<37$ are possible. These are $2^{2}, 2^{3}, 2^{4}, 2^{5}, 3^{2}$, $3^{3}, 5^{2}$.

For $p_{1}=2, k=2$, we get $p_{2}=11$ and as a second solution $n=2^{2} \cdot 11=44$. The remaining powers do not yield a prime $p_{2}$.

For (2): We have

$$
s=p_{1}+p_{1}^{2}+\ldots+p_{1}^{k}=p_{1}\left(1+p_{1}+p_{1}^{2}+\ldots+p_{1}^{k-2}\right)=39=3 \cdot 13
$$

Since $p_{1}<\left(1+p_{1}+p_{1}^{2}+\ldots+p_{1}^{k-2}\right)$, only $p_{1}=3$ is possible. Indeed, $s=3+9+27=39$. Thus, $n=3^{4}=81$ is the third solution. Therefore, there are three numbers that satisfy the required condition: $n_{1}=44, n_{2}=74, n_{3}=81$.