6. Let $n$ be the smallest positive integer satisfying the following conditions: (1) $n$ is a multiple of 75; (2) $n$ has exactly 75 positive divisors (including 1 and itself). Find $\frac{n}{75}$.
6. Solution: From condition (1), let $n=75 k(k \in \mathbf{N})$.

Since $75=5^{2} \cdot 3$, we have $n=5^{2} \cdot 3 k$.
Let $n=5^{s} \cdot 3^{t} p_{1}^{\sigma_{1}} p_{2}^{a_{2}} \cdots p_{m}^{\alpha_{m}}$, where $p_{i}(i=1,2, \cdots, m)$ are prime numbers, and $a_{i}$ are non-negative integers.
From condition (2), in the prime factorization of $n$, the exponents of the primes should satisfy $(s+1)(t+1)\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{m}\right.$ $+1)=75$.

The exponents are 2, 4, 4 or 2, 24 or 4, 14. To make $n$ the smallest, it is clear that $n=5^{2} \cdot 3^{4} \cdot 2^{4}$. Thus, $k=\frac{n}{75}=3^{3} \cdot 2^{4}=432$.