Number of trailing zeros of N!

Description:

Write a program that will calculate the number of trailing zeros in a factorial of a given number.

N! = 1 * 2 * 3 * ... * N

Be careful 1000! has 2568 digits...

For more info, see: http://mathworld.wolfram.com/Factorial.html

Examples

zeros(6) = 1
# 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720 --> 1 trailing zerozeros(12) = 2
# 12! = 479001600 --> 2 trailing zeros

My solution:

function zeros (n) {
  if(n<5) {
    return 0;
  }
  let i = 1;
  let counter = 0;
  while(Math.pow(5,i) <= n) {
    counter += parseInt(n / Math.pow(5,i));
    i++;
  }
  return counter;
}

Best solutions:

이것도 코드는 이해가 가는데 왜 그렇게 짯는지 잘 이해가 안간다.