The number of zeros in the end of a factorial depends on the number of 10s; i.e. effectively, on the number of 5s (since, $10=5\times 2$).

10! is $1\times 2\times 3\times 4\times \left(5\right)\times 6\times 7\times 8\times 9\times (2\times 5)$. The number of zeros at the end of 10!=2

(This is because the highest power of 5 till 10! is 2 (as there are 2 fives))

Continuing like this, 10!-14!, the highest power of 5 will be 2. The next 5 will be obtained at 15 = (5*3).

Therefore, from 15! To 19! - The highest power of 5 will be 3.

Similarly, from 20!-24! - Highest Power = 4

In 25, we are getting one extra five, as 25=5*5. Therefore, 25! to 29!, we will get the highest power of 5 as 6.
The answer to the question is, therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.