[Tutor] puzzled again by decimal module
Tim Peters
tim.peters at gmail.com
Sat Aug 19 03:07:11 CEST 2006
[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
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