Answer: 432

Small Hint:

Factor $75$ into possible products of numbers one greater than prime exponents

Big Hint:

The exponent pattern $(4,4,2)$ can include the required factors $3$ and $5^2$ while assigning the largest exponents to the smallest primes

Solution:

The multiplicative partitions of $75$ give exponent patterns $(74),$ $(24,2),$ $(14,4),$ and $(4,4,2).$ A number divisible by $75=3\cdot5^2$ needs both primes $3$ and $5,$ so the one-prime pattern is impossible. The smallest candidates from the two-prime patterns are $3^{24}5^2$ and $3^{14}5^4,$ respectively. The smallest three-prime candidate is $$n=2^4\cdot3^4\cdot5^2.$$ It has $(4+1)(4+1)(2+1)=75$ divisors, and each two-prime candidate is larger because $$\frac{3^{24}5^2}{n}=\frac{3^{20}}{16}\gt1$$ and $$\frac{3^{14}5^4}{n}=\frac{3^{10}5^2}{16}\gt1.$$ Therefore $$\frac n{75}=\frac{2^4\cdot3^4\cdot5^2}{3\cdot5^2}=2^4\cdot3^3=432.$$