n! is 1 x 2 x 3 x 4 x 5 x .. x (n -1) x n.
Trailing zeros is the result of multilpy even number and every fifth number.
We have enough even numbers. So we should count the number of “every fifth”.
Additional zeros is given by numbers which are the multiply of more than one 5s (25, 125, 625 …)
In general we have:
We can include negative values intercepting but our method is able to handle this situation because number of trailing zeros of non existing (-n) factorial is 0