A simple method is to first calculate factorial of n, then count trailing 0s in the result (We can count trailing 0s by repeatedly dividing the factorial by 10 till the remainder is 0).
The above method can cause overflow for a slightly bigger numbers as factorial of a number is a big number (See factorial of 20 given in above examples). The idea is to consider prime factors of a factorial n. A trailing zero is always produced by prime factors 2 and 5. If we can count the number of 5s and 2s, our task is done. Consider the following examples.
n = 5: There is one 5 and 3 2s in prime factors of 5! (2 * 2 * 2 * 3 * 5). So count of trailing 0s is 1.
n = 11: There are two 5s and three 2s in prime factors of 11! (2 8 * 34 * 52 * 7). So count of trailing 0s is 2.