Trailing zeros of 100!

[Chris Angelico]

On Sun, Jan 3, 2016 at 3:56 AM, Robin Koch <robin.koch at t-online.de> wrote:
> Am 02.01.2016 um 17:09 schrieb Tony van der Hoff:
>>
>> On 02/01/16 16:57, Robin Koch wrote:
>>>
>>> sum([int(0.2**k*n) for k in range(1, int(log(n, 5))+1)])
>>
>>
>> But did you actually test it?
>
>
> Yes, should work for n >= 1.
>
> Why do you ask?

Your "should work" does not sound good as a response to "actually
test". Normally I would expect the response to be "Yes, and it worked
for me" (maybe with a log of an interactive session). Floating point
can't represent every integer, and above 2**53 you end up able to
represent only those which are multiples of ever-increasing powers of
two; 100! is between 2**524 and 2**525, so any float operations are
going to be rounding off to the nearest 2**471 or thereabouts.
That's... a lot of rounding. That's like trying to calculate whether
pi is rational, but basing your calculations on the approximation
3.14. :)

ChrisA