jmc

The prime factorization of $75=3^1{5}^2=(2+1)(4+1)(4+1)$. For $n$ to have exactly $75$ integral divisors, we need to have $n=p_1^{e_1-1}{p}_2^{e_2-1}\cdots$ such that $e_1{e}_2\cdots =75$. Since $75|n$, two of the prime factors must be $3$ and $5$. To minimize $n$, we can introduce a third prime factor, $2$. Also to minimize $n$, we want $5$, the greatest of all the factors, to be raised to the least power. Therefore, $n=2^4{3}^4{5}^2$ and $\frac{n}{75}=\frac{2^4{3}^4{5}^2}{3\cdot 5^2}=16\cdot 27=\boxed{432}$.