4.40 A natural number greater than 1, if it is exactly equal to the product of its distinct proper divisors (excluding 1 and itself), then it is called "good". Find the sum of the first ten "good" natural numbers.
(5th American Invitational Mathematics Examination, 1987)
【Sol】 Let $k$ be a positive integer, and let
$$
1, \quad d_{1}, \quad d_{2}, \cdots, d_{n-1}, \quad d_{n}, k
$$

be all the divisors of $k$, arranged in increasing order.
From this, we have
$$
1 \cdot k=d_{1} d_{n}=d_{2} \cdot d_{n-1}=\cdots
$$

If $k$ is "good," then $k$ equals the product of all its distinct proper divisors, i.e., $k = d_{1} d_{2} \cdots d_{n}$.
From (1),

we have
$$
\begin{array}{c}
k=d_{1} d_{n} \\
d_{1} d_{n}=d_{1} d_{2} \cdots d_{n-1} d_{n}
\end{array}
$$

Thus,
$$
\begin{array}{c}
d_{n}=d_{2} \cdots d_{n-1} d_{n} . \\
n=2 .
\end{array}
$$

Therefore, all the proper divisors of $k$ are $d_{1} \cdot d_{2}$.
Clearly, $d_{1}$ must be a prime. Otherwise, if $d_{1}$ is composite, then a prime factor $p$ of $d_{1}$ must appear in the above increasing sequence of divisors, and we would have $1 < p < d_{1}$, which is impossible.

Similarly, $d_{2}$ must be a prime or the square of $d_{1}$. Otherwise, if $d_{2}$ is composite and not the square of $d_{1}$, then there cannot be only one prime $d_{1}$ between 1 and $d_{2}$.

Thus, $k$ is either the product of two distinct primes $d_{1}$ and $d_{2}$, $d_{1} d_{2}$, or the cube of a prime, $d_{1}^{3}$.
In this way, we can easily list the first 10 "good" numbers:
$$
6,8,10,14,15,21,22,26,27,33 .
$$

Their sum is 182.