It's "testing" 2^25,000,009 - 1 right now. It can test one factor every 1.3
seconds. AUGH! At that rate it would take 95 *billion* years to trial divide
by all odd numbers under 2^62. Nooo
Don't forget, it's not just odd numbers, you only need to trial divide by
numbers that end in 1, 3,
However, a semi-reasonable task would be to test numbers for factors up to
2^16.
Done.
Pitiful, I know, but a TI could test a single number in 12 hours.
An optimized algorithm will do it in about zero seconds.
B) To Mr. Woltman or Mr. Kurowski - how "useful" would factoring (most
likely
[EMAIL PROTECTED] writes:
So, is this:
(2^p mod f) - 1
Congruent to this:
(2^p -1) mod f
Yes, though be careful about the case of 2^p mod f being 0. The first
will give you -1 and the second is f-1. They are congruent, mod f, of
course, but not identical.
This is doubly
