title: Finding Factorials and Ending zeros of a Factorial Number
body: <p><strong>I have a problem. I want to find factorial of big numbers.</strong></p>
  
  <p>Ex: 1555! = ?.
  195! = ?.</p>
  
  <p>My main problem is that I want to know the exact number of ending 0's of the factorial numbers.</p>
  
  <p>I use the following formula:
    (m!)^n = m! = 2*10^(n-1) + 2^2 * 10^(n-2) + ------- + 2^n.</p>
  
  <p>with this I can solve the other factorials for number of ending 0's like this.</p>
  
  <p>100!= 2*10^1 + 2^2*10^0 = 20+4 = 24</p>
  
  <p>100! has 24 ending 0's as per this calculation.</p>
  
  <p>But, then I got other problem, </p>
  
  <p>Ex: For 95!</p>
  
  <p>i) 95! = (100 - 5)! = 24 - 2*5^(1-1) = 24 - 2 = 22 => 95! has 22 0's.</p>
  
  <p>ii) 95! = (90 + 5)! = 9*(2*10^0) + 2*5^0)= 18+2 = 20 => 95! has 20 0's.</p>
  
  <p>this is my problem. By using the above formula I got two different answers and I am confused, I don't get the perfect answer so please help me to find it.</p>
  
  <p><em><strong>Thank you...</em></strong></p>