5. (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 \( \frac{n}{75} \).
5. Since $75=5 \times 5 \times 3, n$ has exactly 75 divisors, and $n$ is a multiple of 75, so $n=75 k=3 \times 5^{2} k \geqslant 2^{1} \times 3^{1} \times 5^{2}$.

By the minimality of $n$, $\frac{n}{75}=k=2^{1} \times 3^{3}=432$.