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Old 2014-10-31, 00:38   #1
zakary
 
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my numbre with 101 019 520 digits. how can i display it in internet??
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Old 2014-10-31, 00:47   #2
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Just type it into a word document, zip it and post it.
Word documents zip much better than plain text files!
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Old 2014-10-31, 00:48   #3
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Inb4miscmath.
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Old 2014-10-31, 02:44   #4
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Quote:
Originally Posted by zakary View Post
my numbre with 101 019 520 digits. how can i display it in internet??
Easy: just write 2^335,579,581-1. At least, I'm guessing that's your "prime" number (plus or minus 2), since that's a Mersenne number with exactly that many digits (indeed, the only one with a prime exponent in a good range around that size). It's even without known factors (even after I pushed its TF 5 bits (and counting) higher than it was), so your crank score isn't quite as high as it could've been. Congrats on that!

Last fiddled with by Mini-Geek on 2014-10-31 at 02:53
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Old 2014-10-31, 03:21   #5
Batalov
 
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Why not 2^(27!+1)-1 ? It is surely bigger and has no known factors... yet? Must be prime, too, and much bigger!
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Old 2014-10-31, 03:53   #6
LaurV
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Quote:
Originally Posted by Batalov View Post
Why not 2^(27!+1)-1 ? It is surely bigger and has no known factors... yet? Must be prime, too, and much bigger!
because it has far more digits than claimed?
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Old 2014-10-31, 05:19   #7
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Quote:
Originally Posted by LaurV View Post
because it has far more digits than claimed?
Batalov was asking the OP, not the guesser.
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Old 2014-10-31, 06:02   #8
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Quote:
Originally Posted by Batalov View Post
Why not 2^(27!+1)-1 ? It is surely bigger and has no known factors... yet? Must be prime, too, and much bigger!
Out of curiosity ...

Digits in 2^p -1 are easy to calculate as ceiling_(p*log(2))

Is there a way to easily calculate the digits in something like x! ?

if P = 27! +1 it is clear that 2^p is much, much, much bigger than 100 M digits. I would like to be able to tell how many digits...

Last fiddled with by Primeinator on 2014-10-31 at 06:03
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Old 2014-10-31, 06:36   #9
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Code:
(23:33) gp > \p 30
   realprecision = 38 significant digits (30 digits displayed)
(23:34) gp > (27!+1)*log(2)/log(10)
%4 = 3277876323445093851079375875.23
=>
It has 3277876323445093851079375876 decimal digits, but it has the usual properties of Mersenne's. So the factors are of 2kp+1 form, and \pm 1 (mod 8). I haven't found a factor, but it doesn't mean someone couldn't find one. It could be fun!

P.S. And yes, I was suggesting a parallel to the OP's logic: "No factors are known, therefore it is surely a prime".

P.P.S. There is a Stirling formula for a pretty good estimate of the value of x!

Last fiddled with by Batalov on 2014-10-31 at 06:42
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Old 2014-10-31, 06:40   #10
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3277876323445093851079375876 decimal digits.

(an idiot throws a stone in the lake and 100 sane scientists try to get it out... (romanian proverb))

[edit, grrr, I am always a step behind, hehe... ]
[edit 2: and yes, by idiot I was referring to OP, and not to Serge ]
[edit 3: 2^M48-1 may be larger... or M48#+1 ? (we had this discussion at least ten times on this forum, now other people will jump in and say larger possible primes)]
[edit 4: scrap M48#, we can not construct it, as we don't know all primes up to it ]

Last fiddled with by LaurV on 2014-10-31 at 06:49
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Old 2014-10-31, 06:55   #11
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Quote:
Originally Posted by Mini-Geek View Post
Easy: just write 2^335,579,581-1. At least, I'm guessing that's your "prime" number (plus or minus 2), since that's a Mersenne number with exactly that many digits (indeed, the only one with a prime exponent in a good range around that size). It's even without known factors (even after I pushed its TF 5 bits (and counting) higher than it was), so your crank score isn't quite as high as it could've been. Congrats on that!
If you haven't started a P-1 on it, I might be able to do that on one of my borg.
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