New Largest Prime Found: Over 7 Million Digits 305
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."
Not in this case... (Score:3, Insightful)
hmmm (Score:2, Insightful)
Personally I think someone should work this out on paper. Any volunteers/nominations?
Re:What kind of data structures are used? (Score:1, Insightful)
Re:What kind of data structures are used? (Score:1, Insightful)
For a real free bignum library, try LibTomMath [libtomcrypt.org]. It was written by a guy named Tom.
Yeah, but (Score:1, Insightful)
Re:I hate to be a pushover... (Score:3, Insightful)
Verification (Score:3, Insightful)
Re:Not in this case... (Score:5, Insightful)
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
Re:Not in this case... (Score:5, Insightful)
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
Want a simple proof? (Score:5, Insightful)
= (-1)^(odd number)+1 [mod 3]
= -1 + 1 [mod 3]
= 0 [mod 3]
Slashdot Earns It's Name (Score:3, Insightful)
I love it!
Re:I hate to be a pushover... (Score:4, Insightful)
Re:Persuit of uselessness != profit (Score:3, Insightful)
I ask because your tirade, although vigorous and interesting, is entirely unrelated to my post.
Since you seem to be articulate and well read I'm giving you the benefit of the doubt and assuming that you have some sort of agenda...
What is it exactly?
Re:I hate to be a pushover... (Score:4, Insightful)