6. (10 points) Given that the natural number $n$ has 10 divisors, $2 n$ has 20 divisors, and $3 n$ has 15 divisors, then $6 n$ has $\qquad$ divisors.
【Analysis】$n$ has 10 divisors, while $2n$ has 20 divisors. According to the divisor sum theorem, we know that the factorization of $n$ does not contain 2. If $3n$ has 15 divisors and assuming the factorization of $3n$ does not contain 3, then the number of divisors of $3n$ should be $(1+1) \times 10 = 20$. Therefore, the factorization of $n$ contains one 3. For $6n$, which is $2 \times 3 \times n$, we can use the divisor sum theorem to find the number of divisors.

【Solution】According to the analysis, $n$ has 10 divisors, and $2n$ has 20 divisors. By the divisor sum theorem, and since $\frac{20}{10} = \frac{1+1}{0+1}$, the prime factorization of $n$ contains 0 twos. Let $n = 3^a m^x$. Since $\frac{15}{10} = \frac{2+1}{1+1}$, the prime factorization of $n$ contains one 3. According to the divisor sum theorem, the number of divisors of $n$ is: $(a+1)(x+1) = 10$,

Solving this, we get: $a = 1, x = 4$, so $n = 3 \times m^4$;
Thus, $6n = 2 \times 3 \times n = 2 \times 3 \times 3 \times m^4 = 2 \times 3^2 \times m^4$,
The number of divisors is: $(1+1) \times (2+1) \times (4+1) = 2 \times 3 \times 5 = 30$,
Therefore, the answer is: 30.