For a positive integer n, if 5^n is a factor of 20!, but 5^(n+1) is no

I don't get how the question asks the number of 5 in 20!.. could you elaborate?

There are 2 questions in the stem, let's break them down real quick:
1) 20!/5^n = integer
2) 20!/5^n+1 ≠ integer

Q1: If 20!/5^n = integer, then n could max be 4. If you prime factorize the 5's in 20! (20, 15, 10 and 5) you will end up in 5^4, this is the maximum value where 20!/5^4 = integer

Q2: We already determined that for 20!/5^n to be an integer, n can max be 4. If n = 1, then this will still be an integer, same where n = 2 or 3, but when n = 4 the result will be 20!/5^5. In Q1 we figured out that the maximum value for n has to be 4, so that 20!/5^4 = integer. If n = 4 in this question, we get 20!/5^5 which will not be an integer

Conclusion: for n to be an integer in Q1 and not an integer in Q2 n has to be 4

We have to find the greatest possible value of integer n such that:

20!/5^n = integer

The expression above is an integer if n is not greater than the total number of 5s in the prime factored form of 20!.

We can quickly determine the total number of 5s with the following technique:

20/5 = 4

Therefore, the maximum value of integer n is 4.

Answer: A