4. Find all natural numbers that have exactly six divisors, whose sum is 3500.
Solution. If the natural number $n$ has exactly six divisors, then $n=p^{5}$ or $n=q^{2} r$ where $p, q$ and $r$ are prime numbers.

(i) Let $n=p^{5}$. Then $1+p+p^{2}+p^{3}+p^{4}+p^{5}=3500$, that is,

$$
p\left(1+p+p^{2}+p^{3}+p^{4}\right)=3499
$$

Since 3499 is a prime number, it follows that this case is not possible.

(ii) Let $n=q^{2} r$. Then $1+q+q^{2}+r+r q+r^{2} q=3500$, that is,

$$
\left(1+q+q^{2}\right)(1+r)=3500=2^{2} \cdot 5^{3} \cdot 7
$$

The number $1+q+q^{2}=1+q(1+q)$ is odd and not divisible by $5\left(1+q+q^{2}\right.$ can give remainders 1, 2, or 3 when divided by 5). Thus, $1+q+q^{2}=7$. From this, we get $q(q+1)=6=2 \cdot 3$, that is, $q=2$. Further, we get $r=499$ and finally, $n=1996$.

## II year