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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."
Say all you will, but Optimus is still the ultimate prime.
I also reply below your current threshold.
Well we can make a perfect number with it.
Every Mersenne prime gives rise to a perfect number.
To answer your question a little more seriously the number is not much use in itself but like many peices of research the route to the goal often turns out more interesting information than the goal. GIMPS pushes back the bounds on many levels such as highly optimised coding and mathematical DC.
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Notice I said it's an attempt, I didn't say it would work;)
If my answers frighten you, stop asking scary questions.
Who knows, one day you might find yourself struck in the tiger den with multiple doors all marked with Mersenne Primes, and a sign saying, "safe exit thru the door marked with the 41st Mersenne Prime". Yeah, then who is gonna bitch about not memorizing that sucker, huh?
Size does matter :)
In binary: 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111...
:/
Your comment violated the "postercomment" compression filter. Try less whitespace and/or less repetition. Comment aborted.
Sorry
I have discovered a truly marvelous demonstration of a 10 million digit prime which this margin is too narrow to contain.
Which is why I'd hate to be a math student in the 22nd century:
Teacher: Ok class, your homework for tomorrow is to find a Mersenne prime longer than 1,000,000 digits. *By hand*. I don't want to see any computer printouts.
Class: *Groan*
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
Wait, I didn't tell that right.
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
"Are Mersennes really the easiest numbers to prove prime?"
Yes, because of the Lucas Lehmer primality test, which you can google if you want to see the details.
The standard proof of primality involves factoring the number one less than or one greater than the prime. Obviously, the number one greater than 2^p-1 is easily factored, which is the basis of the test.
a,e,i,o,u and sometimes w and y (at be if of up cwm by)
2^(odd number)+1
= (-1)^(odd number)+1 [mod 3]
= -1 + 1 [mod 3]
= 0 [mod 3]