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Old 2003-12-19, 09:34   #12
shu_the_genius
 
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233693433-1
:-)
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Old 2003-12-19, 09:35   #13
shu_the_genius
 
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Thank you wblipp for teaching how to post exponents :-)
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Old 2006-01-17, 14:11   #14
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Hello,

(2^2)-1=3

2^((2^2)-1)-1=(2^3)-1=7

2^(2^((2^2)-1)-1)-1=(2^7)-1=127

(2^127)-1=.... is also a prime

then (2^ ((2^127)-1))-1= also prime? and if so, that's mutch bigger then all the prime's that have been found.

provebale? with binary numbers?
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Old 2006-01-17, 14:50   #15
Mystwalker
 
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Unfortunately, it is currently (and probably the next centuries) not possible to prove primality of MM127 - and attempts to find a factor (and thus prove it composite) yielded no success so far.

Interesting links I've found:
http://mersenneforum.org/showthread.php?t=2850
http://homepages.donobi.net/poke/archive5/0066.html (plus replies)

But there are a lot of people here with more knowledge of this issue, so please don't completely rely solely on me...
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Old 2006-01-18, 00:07   #16
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Besides the monstrous runtime estimates given in the above threads for LL testing a number the size of M(M127), there is an even more fundamental problem: how to store numbers that large. M(M127) is a number having 2127 or roughly 1040 bits. Assume we had some magical way of using just a single hydrogen atom to store each bit of such a number (note that Helium would be safer if there's any oxygen nearby). Thus we would need roughly (1014 times Avogadro's number) hydrogen atoms, or roughly 1014 grams of hydrogen to store our data, assuming a data-perfect algorithm that needed only as many bits as the number being tested. That is roughly 1000 cubic kilometers of hydrogen at standard temperature and pressure - roughly equivalent to the volume of atmosphere overlying a small country - or around 1 cubic kilometer in more-compact supercooled-liquid form. Moreover, we'd need a way to coherently and quickly manipulate all the atoms in that volume in parallel. And even if we could do this amazing feat, the large size of this data reservoir would set fundamental limits on how quickly we could compute using it - how long does it take light to cross a spherical reservoir containing 1 cubic kilometer of ultracold liquid hydrogen? The answer is, on the order of a microsecond (thanks, alpertron ;), which limits our maximum operating frequency to around 1MHz, which is in fact much slower than today's microchips!

Thus, in order to complete the computation in less than the life of the universe (based on the time it takes low-mass stars to burn out), our hypothetical computer would have to be many orders of magnitude smaller than it could possibly be simply to store the data. It appears one is trapped between Scylla and Charybdis (the "rock and whirlpool-slash-hard-place" of Greek mythology) - the only way to be fast enough is to be much too small. Call me a pessimist, but I think I'll stick to trial-factoring - there, storing the data (which need only be the size of the factor candidates being tried) is no problem, and neither is massive parallelism (many machines can work on the problem, each on its own range without interacting with the others, except to get new ranges to test.)

(I've gone up to around 176 bits using my own factoring code - no factors yet.)

Last fiddled with by ewmayer on 2006-01-18 at 01:23 Reason: Since this topic is closely related to the more general issue of whether 2[sup]{Some Mersenne prime}[/sup]-1 is prime, I've merged it into that thread.
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Old 2006-01-18, 21:52   #17
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Does anyone know the speed difference between factoring MM127 with MFAC and using Factor_4 on M(170141183460469231731687303715884105727)? (the long version of MM127)
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Old 2006-01-20, 00:20   #18
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Ok, I don't know if this will require some kind of high level math knowledge to understand the answer, but I have a question. Why and how is it possible to find a factor of a number when you can't calculate the exact number itself without filling the universe or creating an impossible circumstance?
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Old 2006-01-20, 03:23   #19
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Quote:
Originally Posted by Jwb52z
Ok, I don't know if this will require some kind of high level math knowledge to understand the answer, but I have a question. Why and how is it possible to find a factor of a number when you can't calculate the exact number itself without filling the universe or creating an impossible circumstance?
Modular arithemetic. It allows you to calculate the remainder of an expression using arithemetic only about twice the size of the divisor. For example, to find the remainder of 10^1000000 after dividing by 3, we observe that 10 divided by 3 leaves 1, so the answer is the same as 1^1000000 = 1. In particular, we now know that 3 divides 10^1000000+2, never using numbers larger than 10 in the calculation.
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Old 2006-01-20, 11:34   #20
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As an interesting application of modular arithmetic you can see on my site the factorization of number near googolplex and factorization of number near googolplexplex, where googolplex = 10^(10^100) and googolplexplex = 10^googolplex.

Of course you can't expect to completely factor these numbers, but at least some factors can be found and show that most of these numbers are composite.
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Old 2006-01-20, 21:03   #21
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Default Question about the Googol program

Hi Dario,

I'm interested in your Googol program but I'm a bit confused.

At the top, it says for the numbers involved (the range of googolplex +/- 1000) that:

They have no other prime factors less than 330 x 10^12

but the program at the bottom says:

This version supports numbers up to 2 x 10^14.

So that implies to me that the code would have to be modified to find any new factors.

=======================================================

Also, is there a similar program on your site to find factors of numbers near googolplexplex?

Thanks,
Grandpa

Last fiddled with by grandpascorpion on 2006-01-20 at 21:03
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Old 2006-01-21, 19:09   #22
ewmayer
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Quote:
Originally Posted by wblipp
Modular arithemetic. It allows you to calculate the remainder of an expression using arithemetic only about twice the size of the divisor. For example, to find the remainder of 10^1000000 after dividing by 3, we observe that 10 divided by 3 leaves 1, so the answer is the same as 1^1000000 = 1. In particular, we now know that 3 divides 10^1000000+2, never using numbers larger than 10 in the calculation.
Note that this only works if the number to be factored has some convenient algebraic form involving only small integers, e.g. N = ab+c, where a,b,c are all much smaller than N in magnitude.
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