Zhendarov R.G.

Find all natural numbers that have exactly six divisors, the sum of which is equal to 3500.
If a number $n$ has six divisors, then $n=p^{5}$ ( $p$ - prime) or $n=p^{2} q$, where $p$ and $q$ - are different prime numbers.

In the first case $1+p+p^{2}+p^{3}+p^{4}+p^{5}=3500, p\left(p+p^{2}+p^{3}+p^{4}\right)=3499$. The number 3499 is not divisible by $2,3,5$ and 7, so $p>10$, but in this case

$p\left(p+p^{2}+p^{3}+p^{4}\right)>3499$. Therefore, this equation has no solutions in prime numbers.

In the second case $1+p+p^{2}+q+p q+p^{2} q=3500$, that is $\left(1+p+p^{2}\right)(1+q)=5^{3} \cdot 7 \cdot 4$. The first factor is odd and not divisible by 5. (To verify this, it is sufficient to check this statement for the corresponding remainders.) Hence, considering that $1+p+p^{2}>1$, we have $1+p+p^{2}=7$. Therefore, $p=2, q=499$. The numbers 2 and $499-$ are prime. The required number $n=2^{2 \cdot 499}=1996$.

## Answer

$n=1996$.

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