Slashdot Mirror
Largest Prime Number Discovered – With More Than 23m Digits (mersenne.org)
chalsall writes: Persistence pays off. Jonathan Pace, a GIMPS volunteer for over 14 years, discovered the 50th known Mersenne prime, 2^77,232,917 -- 1 on December 26, 2017. The prime number is calculated by multiplying together 77,232,917 twos, and then subtracting one. It weighs in at 23,249,425 digits, becoming the largest prime number known to mankind. It bests the previous record prime, also discovered by GIMPS, by 910,807 digits. You can read a little more in the press release.
Just think how big a prime PHOTOSHOPS could find!
"At present there are few practical uses for this new large prime, prompting some to ask "why search for these large primes"? Those same doubts existed a few decades ago until important cryptography algorithms were developed based on prime numbers. For seven more good reasons to search for large prime numbers, see here.
The more prime numbers that are discovered, the more likely we are to be able to discover a pattern within an arbitrary base number set. The larger numbers are useful because we also want to make sure that the entire range is consistent, or in other words that any pattern, or lack of pattern, is the same across the entire set of numbers. There is always a benefit to trying to find patterns in number theory -- it's one of the coolest and most interesting fields in pure mathematics.
The Spanish Inquisition of Psychometrics; Burning all the heretics.
Subject says it all.
think of all the dogecoin they could have mined!
Primarily for the fun of it. There are some specific uses of large Mersenne primes in the Mersenne twister algorithm for generating pseudorandom numbers https://en.wikipedia.org/wiki/Mersenne_Twister, but in practice much, much smaller Mersenne primes are perfectly fine for that use, and indeed are much more practical. There are people who whenever you talk about large primes will claim they are useful for crypto, but that's not generally the case. The primes are too big for practical Diffie-Hellman (and there are specific reasons one might want to avoid using them for that), and they are not random primes in any sense so using them for any form of RSA would be really silly. That said, there's at least one mildly fun cryptographic algorithm whose proof of correctness relies on there being infinitely many Mersenne primes http://www.di.ens.fr/~vergnaud/algo0910/Locally.pdf, but no one has to my knowledge actually tried to implement the algorithm in that paper.
Exercise for the interested maths nerd: Prove that if q is any even number, then 2^q - 1 is divisible by 3.
sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});
That's amazing! I've got the same combination on my luggage!
rewriting history since 2109
FirehoseFavorites is purely user voted content. Something new we are testing. Requires zero editor input to make it to the front page, just user votes from the firehose.
FirehoseFavorites is purely user voted content. Something new we are testing. Requires zero editor input to make it to the front page, just user votes from the firehose.
Ah, that makes sense - thank you for the explanation.
#DeleteChrome
Unleash the bots!
Not a rigorous proof, but here's my favorite explanation:
for any positive integer k, the binary representation of 2^k-1 consists of k 1's. If k is even, this is an even number of 1's lined up together. Since 3 is 11 in binary, you can divide 2^k-1 by 3 and get a quotient of the form 10101..01.
e.g. 2^10 = 1111111111=11(101010101)
To clarify, these types of primes aren't useful for cryptography as they are much too large and not 'typical' primes.
From a theoretical perspective they are quite interesting: they are in bijective correspondence with perfect numbers and no one knows whether there are infinitely many. For all we know, this one could theoretically be the last.
"What lies behind us, and what lies before us are tiny matters compared to what lies within us." Ralph Waldo Emerson
Exercise for the interested maths nerd: Prove that if q is any even number, then 2^q - 1 is divisible by 3.
Because q is even, we can write 2^q - 1 = [2^r -+1] * [2^r - 1], where 2r = q and r is a (positive) integer.
Now consider the numbers 2^r - 1, 2^r, and 2^r + 1. One of these numbers must be divisible by 3. Clearly it is not 2^r because it only has factors that are powers of 2. Therefore either 2^r - 1 or 2^r + 1 must be divisible by 3. QED.
If it weren't for deadlines, nothing would be late.
OK people, we're done here! We found the last prime! Time to shut it down! You don't have to go home, but you can't stay here!
OK, that was a joke but we can still be clear. He was talking about the last perfect number. There is an infinite number of primes. That proof is pretty simple.
1. Assume there is a limited number of primes. Given the list of all the prime numbers
2. Multiply them all together and add 1.
The new number you get is can not divisible by any of the prime numbers in your list (e.g. if you divide the number by 2, you have a reminder of 1, if you divide the number by 3, you have a remainder of 1, if you divide the number by 5, you have a remainder of 1...)
So there must be at least one number not on your list which invalidates the given statement.
IMO, the main practical advantage is that checking for MP is fast, so they're easier to search for. Working with MP number fields is fast too (owing to close-to-pow2 property). so they find use whenever any large prime would do, it's just a nice to have MP in there.
A typical practical usage is PRNGs of girgantual periods (the period is typically the MP itself, or multiple of thereof) for HPC number crunching. The perfect number property indeed is a nice bonus in there, as it often leads to better k-distribution of the permutation.
http://maths-people.anu.edu.au...