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2010-11-12, 20:18   #45
CRGreathouse

Aug 2006

2·2,969 Posts

Quote:
 Originally Posted by science_man_88 23 and 89 in this case.
If you want us to comment on more than just this case, you'll have to give us more information. If not, that's cool too.

2010-11-12, 20:22   #46
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by CRGreathouse If you want us to comment on more than just this case, you'll have to give us more information. If not, that's cool too.
Code:
 (11:39) gp > sumdigits(47)+sumdigits(178481)
%69 = 4
(16:19) gp > sumdigits(47*178481)
%70 = 4
in this case it works without the outer sumdigits in the first expression because both sumdigits are below 5 so it can't add up above 9.

 2010-11-12, 20:24 #47 science_man_88     "Forget I exist" Jul 2009 Dumbassville 20B116 Posts okay found a bad exception 2^29-1 because it uses 3 but if we can reduce it to 2 maybe it still works this is confirmed as a good change lol. Last fiddled with by science_man_88 on 2010-11-12 at 20:28
 2010-11-12, 20:33 #48 science_man_88     "Forget I exist" Jul 2009 Dumbassville 8,369 Posts okay verified exception = 2^37-1 so how to compensate for exceptions if not to many.0 maybe it works for sumdigits(2^p-1) = 4 the hard part is adapting to the case when it's 1. Last fiddled with by science_man_88 on 2010-11-12 at 20:37
2010-11-12, 20:41   #49
axn

Jun 2003

112558 Posts

Quote:
 Originally Posted by science_man_88 Code:  (11:39) gp > sumdigits(47)*sumdigits(178481) %69 = 4 (16:19) gp > sumdigits(47*178481) %70 = 4 in this case it works without the outer sumdigits in the first expression because both sumdigits are below 5 so it can't add up above 9.
See the highlighted part. That is the correct relation. This follows from the fact that sumdigits(x) is equivalent to x%9.

2010-11-12, 20:44   #50
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000101100012 Posts

Quote:
 Originally Posted by axn See the highlighted part. That is the correct relation. This follows from the fact that sumdigits(x) is equivalent to x%9.
it works for addition for sumdigits(x*y)=4 lol

2010-11-12, 20:46   #51
axn

Jun 2003

7·683 Posts

Quote:
 Originally Posted by science_man_88 it works for addition for sumdigits(x*y)=4 lol
not really. try the trivial case 1*4.

2010-11-12, 20:52   #52
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by axn not really. try the trivial case 1*4.
i mean't the mersenne prime factors and so far I've got it working for the first 3 of sumdigits(2^p-1)=4

 2010-11-12, 21:40 #53 science_man_88     "Forget I exist" Jul 2009 Dumbassville 836910 Posts okay i found the exception at 101 point noted. I'll see if either of the idea's work if one does maybe we can limit even more.
 2010-11-12, 23:32 #54 science_man_88     "Forget I exist" Jul 2009 Dumbassville 8,369 Posts Code: 123456789 246813579 369369369 483726159 516273849 639639639 753186429 876543219 999999999 this is the 2 way multiplication for modulo if I have it correct, the red are ones that can if multiplied by something else or (for some) left alone can be brought to 4 or 1 as sumdigits(x)
 2010-11-13, 02:17 #55 3.14159     May 2010 Prime hunting commission. 24×3×5×7 Posts I think this is a long-winded circular argument again.. Hopefully it doesn't take 25 pages of posts to realize this. Last fiddled with by 3.14159 on 2010-11-13 at 02:18

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