New Largest Prime Found: Over 7 Million Digits
Gilchrist continues "If you want to see the number in written in decimal, Perfectly Scientific, Dr. Crandall's company which developed the FFT algorithm used by GIMPS, makes a poster you can order containing the entire number. It is kind of pricey because accurately printing an over-sized poster in 1-point font is not easy! Makes a cool present for the serious math nut in your family.
For more information, the press release is available.
Congratulations to Josh and every GIMPS contributor for their part in this remarkable find. You can download the client for your chance at finding the next world record prime! A forum for newcomers is available to answer any questions you may have.
GIMPS is closing in on the $100,000 Electronic Frontier Foundation award for the first 10-million-digit prime. The new prime is 72% of the size needed, however an award-winning prime could be mere weeks or as much as few years away - that's the fun of math discoveries, said GIMPS founder George Woltman. The GIMPS participant who discovers the prime will receive $50,000. Charity will get $25,000. The rest will be used primarily to fund more prime discoveries. In May 2000, a previous participant won the foundation's $50,000 award for discovering the first million-digit prime."
Nothing. And we all die, too. Might as well kill yourself right now, avoid all the unnecessary hassle of life. Ya know.
Theorem For any positive odd integer n, 3 divides 2^n+1
Proof We will use the Principal of Mathematical Induction.
Basis When n=1, we have 2^n+1=2^1+1=3. Furthermore, when n=3, we have 2^n+1=2^3+1=9.
Induction Now suppose n is a positive odd integer, and that 3 divides 2^n+1. We will now show that 3 divides 2^(n+2)+1.
Since 3 divides 2^n+1, there exists an integer q such that 2^n+1=3*q
2^(n+2)+1=2^(n+2)+4-3
=2^2*2^n+4-3
=4*(2^n+1)-3
=4*3*q-3
=3*(4*q-1)
=3*r, r=4*q-1
Where r is an integer by the closure properties of multiplication and subtraction.
QED
More directly (without induction):
if T = 2^(2p+1) + 1:
T = 2^(2p+1) - 2 [mod 3]
T = 2(2^2p - 1) [3]
T = 2(4^p - 1) [3]
T = 2(1^p - 1) [3]
T = 0 [3]
qed
2^(odd number)+1
= (-1)^(odd number)+1 [mod 3]
= -1 + 1 [mod 3]
= 0 [mod 3]
This story is perhaps the most pure example of "News for Nerds. Stuff that matters."
:- )
I love it!
They're not useful now, but in another 200 years when we're all carrying around pocket quantum machines it may be useful.
Don't say that out loud on a star trek convention.
Be wary of any facts that confirm your opinion.