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I have convinced myself that this true, however I'm at a loss of where I should start with this proof. Looking at a similar proof with 3 instead of 9, I saw the use of the basis representation theorem, but I'm not 100% comfortable with that, so it was hard for me to follow. Is there a way to do this with mod?

Hippalectryon
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user188693
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    This holds for all $n$, not just three-digit numbers. The proof isn't too hard, and, yes, it does include modular arithmetic. – Edward Jiang Oct 31 '14 at 01:05
  • If you're only concerned about 3-digit numbers, you can just check them one by one, since that's a small, finite set. But you'd be missing out on a proof of the general case. – Robert Soupe Oct 31 '14 at 02:27

2 Answers2

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Hint: $10\equiv1\pmod{9} $.

Clayton
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let $z=a_0+10a_1+100a_2\equiv a_0+a_1+a_2 \mod 9$

Dr. Sonnhard Graubner
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