4. There are $n$ numbers, starting from the second number, each number is 3 more than the one immediately before it, i.e., $4, 7, \cdots, 3n + 1$, and the product of these numbers ends with exactly 32 zeros. Then the minimum value of $n$ is ( ).
(A) 125
(B) 126
(C) 127
(D) 128
4.D.

Since $(1+3 n) \div 5=2+3(n-3) \div 5$, therefore, among these $n$ numbers, only the 3rd, 8th, 13th, 18th, ... numbers are multiples of 5, and they are $5 \times 2, 5 \times 5, 5 \times 8, 5 \times 11, \cdots$. Among them, exactly 1 out of every 5 is a multiple of 25, and exactly 1 out of every 25 is a multiple of 125, .... It is easy to see that
$$
(5 \times 2) \times (5 \times 5) \times (5 \times 8) \times \cdots \times (5 \times 7) = 5^{20} \times A,
$$

where $A$ is not a multiple of 5.
Therefore, $5 \times 77 = 3 n + 1$. Hence $n = 128$.