[CTCI 4th Edition] 19.3

Description: Write an algorithm which computes the number of trailing zeros in n factorial.

Thought: 0 comes from 5 and 2, because in n factorial there will be always enough 2....so every 5 will result in a zero.(25 is 5*5 so result in 2 zeros)

Code:
public static int numZeros(int num) {
    int count = 0;
    for (int i = 5; num / i > 0; i = i * 5) {
        count = count + num / i;
    }
    return count;
}

Explanation:
The first loop we count the first column of 5s in the right side, the second loop count the second column of 5s....

5                  5
10                5
15                5
20                5
25                55
30                5
35                5
40                5
45                5
50                55
55                5
..                  ...
75                55