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Old 2010-11-12, 20:18   #45
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Quote:
Originally Posted by science_man_88 View Post
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.
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Old 2010-11-12, 20:22   #46
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Quote:
Originally Posted by CRGreathouse View Post
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.
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Old 2010-11-12, 20:24   #47
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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
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Old 2010-11-12, 20:33   #48
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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
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Old 2010-11-12, 20:41   #49
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Quote:
Originally Posted by science_man_88 View Post
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.
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Old 2010-11-12, 20:44   #50
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Quote:
Originally Posted by axn View Post
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
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Old 2010-11-12, 20:46   #51
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Quote:
Originally Posted by science_man_88 View Post
it works for addition for sumdigits(x*y)=4 lol
not really. try the trivial case 1*4.
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Old 2010-11-12, 20:52   #52
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Quote:
Originally Posted by axn View Post
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
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Old 2010-11-12, 21:40   #53
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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.
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Old 2010-11-12, 23:32   #54
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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)
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Old 2010-11-13, 02:17   #55
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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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