3. A natural number greater than 1, if it is exactly equal to the product of its distinct proper divisors (factors excluding 1 and itself), then it is called "good". Find the sum of the first ten "good" natural numbers.
3. Let $k$ be a positive integer, and let $1, d_{1}, d_{2}, \cdots$, $d_{n_{-1}}, d_{n}, k$ be all its divisors, arranged in increasing order $\Rightarrow 1 \cdot k=d_{1} \cdot d_{n}=d_{2} \cdot d_{n_{-1}}=\cdots$. If $k$ is "good," then by definition, these products are also equal to $d_{1} \cdot d_{2} \cdots \cdots d_{n-1} \cdot d_{n}$. Thus, $n$ can only be 2 and $d_{1}$ must be a prime number (otherwise, the prime factors of $d_{1}$ would appear between 1 and $d_{1}$ in the sequence). Similarly, $d_{2}$ must be a prime number or the square of $d_{1}$ (otherwise, there would not be only one prime factor $d_{1}$ between 1 and $d_{2}$). Therefore, $k$ is either the product of two different primes or the product of a prime and its square, i.e., the cube of a prime. The first 10 good numbers are easily listed as: $6, 8, 10, 14, 15$, $21, 22, 26, 27$ and 33, and their sum is 182.