🚀 go-pugleaf

> >The Crow wrote:
> >>
> >> Here is what I am trying to do, I want to find the last NON-ZERO
> >> digit of a given factorial.  For instance,
> >>
> >> 5! = 120
> >>
> >> so the last non-zero digit is 2.  I want to be able to do this 
> >> up to 1000. Problem is, no matter how huge of a data-type you use,
> >> there isn't any way for the computer to handle 1000!, it is
> >> however possible to find the last non-zero digit somehow.  One
> >> thing I have tried is as I went through multiplying the series of
> >> numbers in the factorial (5 * 4 * 3 * 2) I would remove all the
> >> trailing ZEROS, I got this to work up to 789, but it wont
> >> work with 1000 and i am not really sure why.  If anyone has a
> >> clue how I can do this let me know.
> >

Believe it or not, you do not need a brute force calculation to find
the answer. Simply look at it another way...

How do you add a zero to a number? answer, multiply it by 10, which
means that broken down to primes, multiple it by 2 and 5.

Well, you get a two everytime you multiply an even number, and you
get a five every time it is divisiable by 5. Since, there is more even
numbers than every 5, then you can say "the number of zeros at the
end is the number of fives, when the number is broken down to primes"

Obvious, you get a 5 at every 5, so there are (1000/5) = 200 obvious
fives. You also get a extra 5 at every 25 so there are (1000/25) = 40
fives from that. Well, you get the pattern...

	(1000/5) = 200
	(1000/25) = 40
	(1000/125) = 8
	(1000/625) = 1
	---------------
		    249 fives or 249 trailing zeros

This will of course break when it hits 3125! which can be solved
by adding another term.