Finding Factorials and Ending 0's for High Values.

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Hello friends,
I have a problem. I want to find factorial of big numbers.
Ex: 1555! = ?.
195! = ?.
My main problem is that I want to know the exact number of ending 0's of the factorial numbers.

I use the following formula:
(m!)^n = m! = 2*10^(n-1) + 2^2 * 10^(n-2) + ------- + 2^n.

with this I can solve the other factorials for number of ending 0's like this.

100!= 2*10^1 + 2^2*10^0 = 20+4 = 24
100! has 24 ending 0's as per this calculation.

But, then I got other problem,
Ex: For 95!
i) 95! = (100 - 5)! = 24 - 2*5^(1-1) = 24 - 2 = 22 => 95! has 22 0's.

ii) 95! = (90 + 5)! = 9*(2*10^0) + 2*5^0)= 18+2 = 20 => 95! has 20 0's.

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.

Thank you...

Posted 6-Feb-13 19:40pm

VISH_a_CODE

Updated 7-Feb-13 22:35pm

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Sergey Alexandrovich Kryukov · Accepted Answer · 2013-02-06T19:45:00

Solution 1

In C, this is not so easy. And I don't know good available library; maybe someone else will point out one for you. You can find some ideas here: http://stackoverflow.com/questions/3340511/what-is-the-simplest-way-of-implementing-bigint-in-c[^].

—SA

Posted 6-Feb-13 19:45pm

Sergey Alexandrovich Kryukov

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VISH_a_CODE 7-Feb-13 2:06am

Thanks Sir, I am trying your hint... :)

Sergey Alexandrovich Kryukov 7-Feb-13 8:40am

You are very welcome.
Good luck, call again.
—SA