cs j math Trailing Zeroes

/*
Write a program that calculates with how many zeroes the factorial of a given number N has at its end.
Your program should work well for very big numbers, e.g. N = 100000.

5^1 :  4617 ÷ 5 = 923.4, so I get 923 factors of 5
5^2 :  4617 ÷ 25 = 184.68, so I get 184 additional factors of 5
5^3 :  4617 ÷ 125 = 36.936, so I get 36 additional factors of 5
5^4 :  4617 ÷ 625 = 7.3872, so I get 7 additional factors of 5
5^5 :  4617 ÷ 3125 = 1.47744, so I get 1 more factor of 5
5^6 :  4617 ÷ 15625 = 0.295488, which is less than 1, so I stop here.
Then 4617! has 923 + 184 + 36 + 7 + 1 = 1151 trailing zeroes.
http://www.purplemath.com/modules/factzero.htm
 */

using System;

class TrailingZeroes
{
    static void Main()
    {
        int  number = int.Parse(Console.ReadLine());
        int count = 1, sumZeroes = 0, truncated = 0;

        do
        {
             truncated = (int)Math.Truncate(number / Math.Pow(5, count));
             sumZeroes += truncated;
             count++;
        }
        while ((truncated) >= 1);

        Console.WriteLine(sumZeroes);
    }
}