Roger Hui wrote:
> Perhaps Python extended precision numbers are represented
> in base 2? That would explain why (2^k)^n is fast but 3^n
> is slow.
>
>
>From the Python documentation:
Long integers
These represent numbers in an unlimited range, subject to available
(virtual) memory only. For the purpose of shift and mask operations, a
binary representation is assumed, and negative numbers are represented
in a variant of 2's complement which gives the illusion of an infinite
string of sign bits extending to the left.
>
> How do you compute the last 10 decimal digits of 2^10^100x
> in Python? The direct phrase
>
> x = (2 ** (10 ** 100)) % (10 ** 10)
>
> probably will not work.
>
>
It's been running for over 30 minutes here.
Best wishes,
John
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