[Tim Peters]
> You would in this case, and that would be wrong. In fp you'd get an
> approximation to the exact n * (1./5 + 1./5**2 + ...) == n/4. (use
> the rule for the sum of an infinite geometric series). For example,
> that way you'd compute that 4! == 24 has 4/4 == 1 trailing zero,
> inste
Tim Peters wrote:
> [Dick Moores, computes 100 factorial as
>
> 9332621544394415268169923885626670049071596826438162146859296389521753229915608941463976156518286253697920827223758251185210916864
>
> but worries about all the trailing zeros]
>
>> Yes, I'm sure you a
>> 9332621544394415268169923885626670049071596826438162146859296389521753229915608941463976156518286253697920827223758251185210916864
>>> Still not exactly correct! I'm bewildered.
>>>
>> The results look the same to me
>> why do you think they're not correct?
>> what is
