[Tutor] puzzled again by decimal module

[Christian Tschabuschnig]

Tim Peters wrote:
> [Dick Moores, computes 100 factorial as
>  93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
> 
>  but worries about all the trailing zeros]
> 
>> <BLUSH> Yes, I'm sure you are. I'd forgotten about all those factors
>> of 100! that end in zero (10, 20, 30, ..., 100).
> 
> And others, like 2 and 5 whose product is 10, or 4 and 25 whose product is 100.
> 
> For a fun :-) exercise, prove that the number of trailing zeroes in n!
> is the sum, from i = 1 to infinity, of n // 5**i (of course as soon as
> you reach a value of i such that n < 5**i, the quotient is 0 at that i
> and forever after).
> 
> In this case,
> 
> 100 // 5 + 100 // 25 + 100 // 125 + ... =
> 
> 20 + 4 + 0 + ... =
> 
> 24

you should do that with floating-point, so that the quotient never get's
zero and the "i=1 to infinity" makes sense. This way you (might) get 25
which is correct; not 24.

but i'm just guessing - i have no way to prove it ;-)