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 Showing results 1 to 3 of 3 Search took 0.01 seconds. Search: Posts Made By: S485122
 Forum: Conjectures 'R Us 2009-01-12, 06:22 Replies: 73 Views: 10,242 Posted By S485122 1 == 8 == -6 == 15 == -13 mod 7 is the same as... 1 == 8 == -6 == 15 == -13 mod 7 is the same as saying 1, 8, -6, 15 and -13 are multiples of 7 plus 1. If we compute (7+1)^m = 7^m+m*7^(m-1)*1+ ... +m*7*1^(m-1)+1^m all terms except 1^m contain a...
 Forum: Conjectures 'R Us 2009-01-11, 17:31 Replies: 73 Views: 10,242 Posted By S485122 2^(3*m)=(2^3)^m (2^3)^m=8^m 8^m=(7+1)^m ... 2^(3*m)=(2^3)^m (2^3)^m=8^m 8^m=(7+1)^m (7+1)^m==(0+1)^m mod 7 1^m=1 so 2^(3*m)==1 mod 7 Jacob
 Forum: Conjectures 'R Us 2009-01-10, 14:13 Replies: 73 Views: 10,242 Posted By S485122 n==1 mod 3. n=3*m+1. ... n==1 mod 3. n=3*m+1. 123*2^n-1=123*2^(3*m+1)-1=123*2*2^(3*m)-1 2^3==1 mod 7 => 2^(3*m)=(2^3)^m==1 mod 7 123*2*2^(3*m)-1==246*1-1 mod 7 245==0 mod 7 Use the same path for the other case. ...
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