4.9 $n$ 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 $\frac{n}{75}$.
(8th American Invitational Mathematics Examination, 1990)
[Solution] From condition (1), let $n=75 k(k \in N)$.
Since
$$
75=5^{2} \cdot 3,
$$

we have
$$
n=5^{2} \cdot 3 k \text {. }
$$

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

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