## The challenge

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…

### Examples

 ``````1 2 3 4 5 `````` ``````zeros(6) = 1 // 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720 --> 1 trailing zero zeros(12) = 2 // 12! = 479001600 --> 2 trailing zeros ``````

## The solution in Java code

Option 1:

 ``````1 2 3 4 5 6 7 8 9 `````` ``````public class Solution { public static int zeros(int n) { int res = 0; for (int i = 5; i <= n; i *= 5) { res += n / i; } return res; } } ``````

Option 2:

 ``````1 2 3 4 5 6 7 `````` ``````public class Solution { public static int zeros(int n) { if(n/5 == 0) return 0; return n/5 + zeros(n/5); } } ``````

Option 3:

 ``````1 2 3 4 5 `````` ``````public class Solution { public static int zeros(final int n) { return (n < 5) ? 0 : (n / 5) + zeros(n / 5); } } ``````

## Test cases to validate our solution

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 `````` ``````import org.junit.Test; import static org.hamcrest.CoreMatchers.*; import static org.junit.Assert.assertThat; public class SolutionTest { @Test public void testZeros() throws Exception { assertThat(Solution.zeros(0), is(0)); assertThat(Solution.zeros(6), is(1)); assertThat(Solution.zeros(14), is(2)); } } ``````