psi20 Posted June 21, 2006 Share Posted June 21, 2006 I have little background in number theory, groups, conditions, and stuff like that. So I got a book called Teach Yourself Mathematical Groups and this is one of the examples. Prove that a necessary and sufficient condition for a number N expressed in denary notation to be divisible by 3 is that the sum of the digits of N is divisible by 3. I see the proof in the book, but I can't get it. It shows what denary notation is, decimal notation written out like 1x10^4 + 2x10^3 ... Let N= a 10^n + b 10^(n-1) + ... + z (The book uses subscripts instead of different letters for the digits a, b, ..., z) The proof says: If 3 divides N, then 3 divides a 10^n + b 10^(n-1) + ... + z . The part I don't understand is 'For all powers of 10, dividing by 3 gives a remainder of 1. So the remainder when N is divided by 3 is a + b + ... z.' How did that work? Link to comment Share on other sites More sharing options...
phyti Posted June 21, 2006 Share Posted June 21, 2006 10^k=(9+1)^k expand this into 9^k+...+1^k all the terms are divisible by 3 except the last hope that helps Link to comment Share on other sites More sharing options...
psi20 Posted June 21, 2006 Author Share Posted June 21, 2006 Ah yeah, now it makes sense, thanks. Link to comment Share on other sites More sharing options...
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