6. Find all positive integers that have 16 divisors, and the sum of all their divisors is equal to 4032.
(2002, Bulgarian Spring Mathematical Competition)
Solution: If a positive integer has 16 divisors, it can only be one of the following cases: $p^{15}, p^{7} q, p^{3} q^{3}, p^{3} q r, p q r s$, where $p, q, r$, and $s$ are all prime numbers. The sum of all divisors are respectively
$$
\begin{array}{l}
1+p+p^{2}+\cdots+p^{15} \\
=(1+p)\left(1+p^{2}\right)\left(1+p^{4}\right)\left(1+p^{8}\right), \\
\left(1+p+p^{2}+\cdots+p^{7}\right)(1+q) \\
=(1+p)\left(1+p^{2}\right)\left(1+p^{4}\right)(1+q), \\
\left(1+p+p^{2}+p^{3}\right)\left(1+q+q^{2}+q^{3}\right) \\
=(1+p)\left(1+p^{2}\right)(1+q)\left(1+q^{2}\right), \\
\left(1+p+p^{2}+p^{3}\right)(1+q)(1+r) \\
=(1+p)\left(1+p^{2}\right)(1+q)(1+r), \\
(1+p)(1+q)(1+r)(1+s) .
\end{array}
$$

Since the sum of these divisors equals $4032=2^{6} \times 3^{2} \times 7$, and the first 4 cases cannot be divisible by $3, 4$ or 7 for all primes $p, p^{2}+1$. Therefore, it can only be
$$
(1+p)(1+q)(1+r)(1+s)=2^{6} \times 3^{2} \times 7 .
$$

Assume without loss of generality that $p2^{4} \times 3^{2} \times 7$.

If $p+1 \geqslant 6$, then $(1+p)(1+q)(1+r)(1+s) \geqslant 6 \times 8 \times 12 \times 14 > 2^{6} \times 3^{2} \times 7$.

In conclusion, there are 4 numbers that satisfy the condition, namely 1722, 1794, 2002, 2145.