40th Mersenne Prime Found

FenwayFrank writes "A release from New Scientist announces that the Great Internet Mersenne Prime Search found another one: 2^20996011 - 1 is prime. Weighing in at 6,320,430 digits (6 megabytes of prime number...), it becomes the world's largest. Slashdot readers may remember then announcement of the 39th Mersenne Prime, a mere 3.5 million digits."

Comments

Time to update all pages

[Anonymous Coward]

The first Mersenne primes are 3, 7, 31, 127, etc. There are only 39 known Mersenne primes.

Well, now it is 40 known Mersenne Primes, and also 6 discovered by the GIMPS: they need to change the front page to reflect this, and also some banners ("the largest 5 Mersenne primes").

I think it's worth noting that GIMPS not only discovers new Mersenne primes, but also is the discoverer of the biggest six known ones.

Well, that's the way it goes...

[Charbal]

Of course, it didn't occur to me to take a look at the Science section before submitting my own copy of this story (which, since it has several other useful links in it, follows):

Michael Shafer, a graduate student at Michigan State University, took time out for a "short victory dance" upon learning his computer had discovered the 40th known Mersenne prime as part of The Great Internet Mersenne Prime Search. The number itself is 2**20996011-1 and when expressed in base 10, has 6,320,430 digits (zipped copy). However, this is not necessarily the 40th Mersenne prime; there could be another between the previous largest known prime (M39=2**13466917-1, also discovered by GIMPS) and this one. Also worth noting is the still-standing USD$100,000 EFF prize for the discover of the first prime of at least 10 million (decimal) digits. GIMPS clients are available for various operating systems as well as information on how GIMPS would distribute the prize. A press release on the achievement is available as well as several articles. Of course, this also means there's a new largest known even perfect number in town.

Re:Fraud

[Anonymous Coward]

Actually, it's a 7.

2^1-1 mod 10 = 1
2^2-1 mod 10 = 3
2^3-1 mod 10 = 7
2^4-1 mod 10 = 5
2^5-1 mod 10 = 1
2^6-1 mod 10 = 3
2^7-1 mod 10 = 7
2^8-1 mod 10 = 5
etc.

20996011 mod 4 = 3 so it's a 7.

6 Megabytes, eh?

[jvmatthe]

If the number is 2^20996011 then it will take 2099602 bits to store it, or 2624501 bytes along with 4 extra bits. Let's just call it 2624502 bytes. Now, 2624502 divided by 1024*1024 (number of bytes I'd say are in a megabyte) is about 2.5. Which is all to say that somewhere around 2.5 megabytes would be required to store this number, not 6 megabytes as the post here claims.

This is all perfectly true, modulo an arithmetic error on my part. :^)

Re:6 Megabytes, eh?

[k98sven]

Yes, 2.5 megabytes is what it'd take to store it in binary form.

If we want it in human-readable form, convert to base-10:
2^20996011 = 10^(20996011*log(2))
20996011*log(2) is about 6,320,000, decimals.
1 decimal = 1 char = one byte = 6 Mb.

Re:Here's something stupid to do.

[Muad'Dave]

You guys are unblocking the file before searching, right? You'll miss instances of your that wrap around eol. Use:

dd if=prime6.txt of=p6.txt cbs=75 conv=block