# Factorial Trailing Zeros

Classical problem. This problem is in fact to count the number of fives from 1 to n as and the number of 2s is obviously greater than that of 5s. So when there is a 5, there will be one more trailing zero in . We have to solve the problem in a time of and we should also take integers that are power of 5 into consider. The best answer is that we divide n by 5 repeatly and added up all the values in the process.

public class FactorialTrailingZeros {
public int trailingZeroes(int n) {
int count = 0;
while (n >= 5) {
n /= 5;
count += n;
}
return count;
}
}