psi20 Posted June 21, 2006 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?
phyti Posted June 21, 2006 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
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