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."

Fraud

[addaon]

The last digit is a four!

Wait, no, it just got slashdotted before it fully loaded...

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.

interesting

[nocomment]

But can it play DVD's? [utm.edu]

Re:interesting

[fatphil]

Holy cow, I can't believe anyone remembers that now!
-tugs forelock-

While I'm here I must show due respect to GIMPS for harnessing hundreds of
thousands of PCs and bulldozing so convincingly through these ranges.

Phil

Re:interesting

[arvindn]

Holy cow, I can't believe anyone remembers that now!

Sure we do! [wikipedia.org]

BTW, Remember me? :)

Finally!

[Rick the Red]

Now I can generate a secure PGP key pair!

Efficient factoring algorithm

[Peter Millerchip]

I think you'll find that factoring the product of two huge primes becomes slightly easier when the attacker knows one of the primes!

Re:Finally!

[SamSim]

Um, not really THAT secure. The whole number (along with the previous best known prime, M39, at "only" 4,000,000 digits, which you'd presumably also be using) is there for anybody to download.

Re:Finally!

[Chris_Jefferson]

Actually, your PGP key would be one of the most insecure :) PGP's strength requires on factoring two primes multiplied together. However we only know one prime this big, so if you key is bigger than this prime, we have a fairly good idea what one of the factors is!

ObTiredOldJoke

[dacarr]

Imagine the beowulf cluster it took to calculate this!

Here's something stupid to do.

[satanami69]

I found my /. ID (209636) shows up 5 times.

I wonder who has the most occurrences.

Re:Here's something stupid to do.

[infornogr]

Statistically, whoever has the forum ID with the lower number of digits. Somehow I think the nine people that would have to duke it out to answer your question won't respond to this post, however.

Re:Here's something stupid to do.

[GigsVT]

I'd guess malda, with UID 1. Maybe if there are other accounts with UID 2-9 that are still active, they might have a chance. :)

Re:Here's something stupid to do.

[KDan]

No. This one [slashdot.org].

Daniel

Re:Here's something stupid to do.

[dacarr]

Eight occurances for me. Film at eleven.

Re:Here's something stupid to do.

[dacarr]

Or rather, if I have a dumbass attack and do '562227'. It's really like 3 or 4.

Re:Here's something stupid to do.

[Carnildo]

11 occurrences of my /. ID, one occurrence of my ICQ number, and fortunately, my Social Security number isn't in there.

Re:Here's something stupid to do.

[PhuCknuT]

597 occurances. booyah.

Re:Here's something stupid to do.

[BhAaD]

BhAaD: 692949 -----6 Times I beat you...ha!

Re:Here's something stupid to do.

[Naikrovek]

mine shows up at least 6133 times:

c:\>cat prime6.txt | grep 667 | wc -l
6133
c:\>

more, if there are multiple occurances per line, but i didn't check.

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

I bet it's not that big

[Anonymous Coward]

Everyone claims their prime is over 6 million digits in length when in reality most are only about 5.5 million digits.

Wow!

[HoldmyCauls]

It's actually 1916 pages in lynx under a 1024x768 framebuffer!!!

Hah, you really thought I actually counted for a second there!

Re:Wow!

[JeffMagnus]

You can claim to be using a console application and then tell us the number of pixels you use. What counts is the rows and columns.

Get it straight leetboy.

40th?

[Tom7]

This is not necessarily the 40th Mersenne Prime, just the 40th that we've found. We still need to prove many ones in between to be composite before we can mark its place as 40th.

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.

Awesome!

[falsification]

This is great news. Once we have the 42nd, we will know the secret of the universe.

Re:Awesome!

[fmlug.org]

Yes but only if we are asking the right question.

math == piracy

[Anonymous Coward]

by a strange coincidence, the 40th prime is also an MP3-encoded audio file of an unreleased Missy Elliot track.

the RIAA is lobbying to have mathematics outlawed due to the $400 billion lost yearly to these illegal primes.

remember kids, learning math makes you a pirate! stick to watching TV and eating delicious Oreo(R) cookies!

Re:not prime

[luckyguesser]

*ahem*
with one exception :)

Re:not prime

[luckyguesser]

okay... corrected version:
Any expression
2^n
where n is a positive integer will be even,
and subtracting 1 will make it odd.

Re:not prime

[jquirke]

That's because ^ is the XOR operator, not the power operator.

So effectively you are doing 2 XOR 3 - 1 = 1 - 1 = 0

Re:not prime

[damien_kane]

To all the other children of the parent, I think the logic that gave him his answer was something like the following:
(2^20996011-1) => 2^(20996011-1)
= 2^(20996010)
So of course, his logic is flawed, but the answer he came up with is correct based on his logic.
(2^20996010)mod 2 = 0
Therefore 2^20996010 is not prime

Maybe this kid should go back to elementary school and be re-taught the proper Order of Operations.

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 [msu.edu], took time out for a "short victory dance" upon learning his computer had discovered the 40th known Mersenne prime [utm.edu] as part of The Great Internet Mersenne Prime Search [mersenne.org]. The number itself is 2**20996011-1 and when expressed in base 10, has 6,320,430 digits [wolfram.com] (zipped copy [wolfram.com]). 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 [eff.org] for the discover of the first prime of at least 10 million (decimal) digits. GIMPS clients are available for various operating systems [mersenne.org] as well as information on how GIMPS would distribute the prize [mersenne.org]. A press release on the achievement [mersenne.org] is available as well as [newscientist.com] several [wolfram.com] articles [utm.edu]. Of course, this also means there's a new largest known even perfect number [utm.edu] in town.

Damn them!

[nytes]

Just when I thought no one could ever guess my password. Now I have to change it again.

Re:Damn them!

[JDWTopGuy]

You too? Crap!

Re:Damn them!

[damien_kane]

That's ok; I'm using the 41st and 42nd Mersienne Primes as my passwords. which is why people have only got to the 40th MP so far...
I can't let the cat out of the bag until I find the 43rd and 44th and change all of my passwords.
You silly people... thinking the 40th was good enough... And didn't you learn that you shouldn't use the same password for everything? Now you see why :P

Re:Damn them!

[nytes]

Hmmm... Mersienne Prime 41, Mersienne Prime 42, Mersienne Prime 43, Mersienne Prime 44.

Hey! That's the combination on my luggage!

If I post the number here...

[Twintop]

...do you think we could /. the /. server?

6 Megabytes??????

[the eric conspiracy]

Since when does it take a byte to represent a base 10 digit?

This number takes more like 21 KB to represent.

Re:6 Megabytes??????

[Ster]

I presume the *text* of the number is ~6 meg, since each digit as a character is a byte (assuming it's ASCII).

-Ster

Re:6 Megabytes??????

[Tom7]

While we're doing decompression, I offer the following 12-byte version:

2^20996011-1

Seriously, though, where did you get the factor of 300 difference? You think it takes less than a 37th of a bit to represent each digit?

Re:6 Megabytes??????

[the eric conspiracy]

You think it takes less than a 37th of a bit to represent each digit?

Pretty simple - binary math - each place of a binary number is a power of two. Since this number is 2^20996011 - 1, it can be represented in binary form as 20996010 1's, thus 21 KB. It's also why you can count to 2^10 - 1 on your fingers if you use each finger as a binary digit. If you try counting in ASCII base 10 digits on your fingers you can only count to 9, or 2^3 + 1.

Re:6 Megabytes??????

[Scott Wood]

And since when are there around 1000 bits in a byte? 20996010 bits is approximately 2.5 MiB.

Re:6 Megabytes??????

[jareds]

21 million bits is not 21 KB, which is 168 thousand bits. It's 20996011 (not 20996010) bits = 2624502 bytes = 2563 kilobytes = 2.503 megabytes.

Re:

[account_deleted]

Comment removed based on user account deletion

Hmm

[quantaman]

By my calcs (could easily be wrong) my UID should show up about 6.3 times, turned out to be 7.

Xbox?

[k98sven]

Quick? Has anyone checked yet to see if it factorizes the X-box public key?

(joke)

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:6 Megabytes, eh?

[jvmatthe]

If we want it in human-readable form...

Right. Let's pick a USEFUL format. Good idea. ;^) Might as well use cuneiform while you're at it.

Re:6 Megabytes, eh?

[cicadia]

I guess you didn't follow the link [mersenne.org] - it's a text document, with the prime in question represented as an ASCII base-10 string. 1 byte per digit. 6+ megabytes total.

It's not the most compact form, to be sure, but it is, as advertised, 6 megabytes of primey goodness.

Re:6 Megabytes, eh?

[jvmatthe]

You got me. I didn't follow the link. Still, pretty offensive to me to see someone store an integer that way. I spend a whole semester teaching my students to store their data in binary form... :^)

Re:6 Megabytes, eh?

[Urkki]

I sure hope you also taught them the significance of byte order... :-)

Re:6 Megabytes, eh?

[evalhalla]

I guess that an even more compact way to store it would be the binary representation of 2099601, since we know how to calculate the number.

Of course, it would be waaaay less impressive :)

Re:6 Megabytes, eh?

[JohnPM]

Yes but it's easily compressible by representing it with the number 20996011 which can be stored using 25 bits. :)

Re:6 Megabytes, eh?

[jvmatthe]

Give that man a cookie. ;^)

And if someone brings up storing numbers in base 20996011...

Re:6 Megabytes, eh?

[JohnPM]

It gets better! 20996011 is necessarily prime as well. So, it's probably something like the 6 millionth prime, so you could store that using
only 22 bits! For further compression we'd probably have to prove the Riemann hypothesis. :)

Re:6 Megabytes, eh?

[Tackhead]

> And if someone brings up storing numbers in base 20996011...

Mersenne #40: "Almost one of your base are belong to me!"

This should be in "YRO"

[stienman]

This should be in "Your Rights Online" since it contains my personal information:

Age, DL#, SSN, and even my IP! (19216801)

Obviously it's a thinly veiled ploy to steal my identity! I'ma gonna have to sue the student who found this. Be sure to check if you're in there! Luckily they don't have my credit card numbers, but I bet the next big prime is going to have all that and more.

Be afraid. Be very, very afraid.

-Adam

Re:So what

[Kevan_moran]

Actualy, I thought that this question, modded down to 0 was a good one,"How will that help anyone?! "

IMAM ( I am a mathematician, well ok I got a Maths degree from Trinity Cambridge ( and I di..dn't like Quicksilver (though I did love Cryptonomicon (and I can write Lisp )))).

But should I run the Seti screen saver, the Climate predication model or this one?

My current vote is the CP model but somehow it feels less sexy

(lastStatement.getSounds()==Statements.SOUNDS_GEE K Y ? statement.replace(lastStatement

New name for this?

[dacarr]

If one googol is 10^100, would this number be about 6 smeagol?

6 million digits can be stored in under 6 megabyte

[notyou2]

It takes 6 megabytes to store 6 million digits if you encode them in ASCII... but this is highly inefficient.

It would be far more efficient to store them in binary coded decimal (BCD, google for it if you don't know what that is), which requires only 4 bits per decimal digit without sacrificing accessibility of the decimal data. In that case it would only take 3 megabytes.

Re:

[account_deleted]

Comment removed based on user account deletion

Re:

[account_deleted]

Comment removed based on user account deletion

Re:6 million digits can be stored in under 6 megab

[notyou2]

Because that involves an actual conversion between bases, which means that re-extracting (to print the digits out in base 10 again, for example) takes a non-trivial amount of time.

On the other hand, the number was probably originally calculated using base-2 arithmetic (I'm assuming), so storing in binary might be more natural anyhow.

Re:6 million digits can be stored in under 6 megab

[fok]

bzip2 that's it! ;D

mugabytes

[epine]

How did 21.0 mebibits turn into six megabytes? I think he meant mugabytes.

Offtopic, Inflammatory, Inappropriate, Illegal, or Offensive comments might be moderated, and messages that are too short might be downrated.

You can quote me

[Alsee]

The obvious mathematical breakthrough would be development of an easy way to factor this number.

-

Re:

[account_deleted]

Comment removed based on user account deletion

Re:You can quote me

[lordjake]

The ability to factor primes. Wow, that would be an amazing breakthrough indeed.

Use Base 2^20996011 - 1 and write it as just 10

[Cerpicio]

Use Base 2^20996011 - 1 and write it as just 10. Seems like that'd be the most efficient way to write it. -- C.

6 Mbytes to represent? I don't think so.

[shrikel]

So you're saying the number takes 6 Mbytes to represent in ASCII? I can do it in 12 bytes:

2^20996011-1

:)

Quit /.-ing their server

[RennieScum]

Mathematica can figure the number quite nicely. Only took about 10 minutes on a P4 with 256MB RAM.

Dunno...can python handle this? bc? (heh)

Subatomic Particles

[Kosher Beef Jerky]

I was informed today that there are, in the universe, approximately 10^450 subatomic particles in the world. A prime number of 6 million digits (6* 10^100000) cannot be possibly demonstrated in any real object. I proceeded to calculate: Apparently the highest resolution for a monitor so far is 3840x2400 (Can anyone find higher?) and it would therefore take about 108506945 monitors of this resolution just to display 1 quadrillion pixels... (10^12). Any thoughts?

Ever closer (to something infinitely far away)

[SamSim]

Another prime number? That's great! Not long now until we've found them all!

Sweet!

[mcp33p4n75]

With this new Mersenne prime one can also form a new largest perfect number. Euclid proved that if 2^k-1 is prime, then 2^(k-1)*(2^k-1) is a perfect number. Anyone care to put this together?

Re:Sweet!

[lordjake]

That new perfect number would have over 12.5 million digits, which would take a while to work out....