20101110, 19:30  #34 
Aug 2006
1011101001001_{2} Posts 

20101110, 20:26  #35 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 

20101111, 01:32  #36 
May 2010
Prime hunting commission.
2^{4}·3·5·7 Posts 

20101111, 04:22  #37 
Aug 2006
3·1,987 Posts 

20101111, 04:42  #38 
May 2010
Prime hunting commission.
3220_{8} Posts 

20101111, 04:55  #39 
Aug 2006
3×1,987 Posts 
The antecedents, of course  you could easily reverse them.
Last fiddled with by CRGreathouse on 20101111 at 05:10 
20101112, 15:44  #40 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
could we add sumdigits(sumdigits(x)+sundigits(y)) = sumdigits(x*y) into the first theory ? I think it works can anyone help prove it ?.

20101112, 19:43  #41 
Aug 2006
3×1,987 Posts 
What are x and y? (Remember: when you're asking for help you need to be specific!)

20101112, 20:01  #42 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
for example 2047 = 23*89
2+4+7=4 mod 9 2+3+8+9 = 4 mod 9 5+8 = 4 mod 9 so the sumdigits for each multiplier 2kp+1 summed together gives the same as the sumdigits for 2^p+1. 
20101112, 20:03  #43 
Aug 2006
5961_{10} Posts 
So... what are x and y?

20101112, 20:03  #44 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

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