4. (1990 AIME Problem 8) \( n \) is the smallest positive integer that satisfies the following conditions:
(1) \( n \) is a multiple of 75.
(2) \( n \) has exactly 75 positive divisors (including 1 and itself).
Find \( \frac{n}{75} \).
4. From condition (1), let $\dot{n}=75 k(k \in \mathbf{N})$.

Since $75=5^{2} \cdot 3$, we have $n=5^{2} \cdot 3 k$.
Let $n=5^{s} \cdot 3^{t} p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{m}^{a_{m}}$, 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
$$
(s+1) \cdot(t+1) \cdot\left(\alpha_{1}+1\right) \cdot \cdots \cdot\left(\alpha_{m}+1\right)=75 \text {. }
$$

Thus, the exponents should be $2,4,4$ or 2,24, or 4,14.
To minimize $n$, it is clearly
$$
n=5^{2} \cdot 3^{4} \cdot 2^{4} \text {. }
$$

Therefore, $k=\frac{n}{75}=3^{3} \cdot 2^{4}=432$.