How many divisors?

$75p^2 =3\times 5^2\times p^2$ If $p\not= 3$ or 5, number of factor= $(1+1)(2+1)(2+1)=18$

If $p=3$ , number of factor= $(3+1)(2+1)=12$

If $p=5$ , number of factor= $(1+1)(4+1)=10$

Since $p$ is prime, the prime factorization of $75p^2$ is $3 \cdot 5^2 \cdot p^2.$

If $p=3,$ then we have $f(3^3\cdot 5^2) = 4 \cdot 3 = 12.$

If $p = 5,$ then we have $f(3 \cdot 5^4) = 2 \cdot 5 = 10.$

For all other primes, we have $f(3 \cdot 5^2 \cdot p^2) = 2\cdot 3 \cdot 3 = 18.$

Therefore, the minimum value is $\boxed{10}$ .

With p=5, $x=5^4 \times 3$ So the number of distinct prime factors of x reaches min. At this p, number of distinct factors of x is: $\boxed{10}$