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Suppose $m = a_k + a_{k -1} + \ldots + a_1 + a_0$.

Does $3$ divide $m$?

If so, how do we prove that?

We know that $3|m \to 3j = a_k + a_{k -1} + \ldots + a_1 + a_0$ for some $j \in \mathbb Z$.

So, then is $j = \frac {a_k + a_{k -1} + \ldots + a_1 + a_0}{3}$? If not, how do we find $j$?

If none of that is correct, how do we go about showing $3|m$?

Erbolat
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    To answer requires further knowledge about the $,a_i.,$ – Bill Dubuque Jul 26 '14 at 15:46
  • @Bill I need more coffee ;-) (I'm shaking my head at myself). Thanks for the query! – amWhy Jul 26 '14 at 15:49
  • @Bill Dubuque, why can't we define $j$ the way it's done in the OP? What law would that violate? Also, how does $a_i$ help us here? – Erbolat Jul 26 '14 at 16:00
  • @Erbolat If $,3\mid m,$ then $, j := m/3,$ is an integer. But you cannot generally deduce that $,3\mid m,$ without knowing more about the $,a_i.,$ Is there some further context? Are they the decimal digits of an integer? – Bill Dubuque Jul 26 '14 at 16:17
  • I am trying to prove that if $n = a_ka_{k -1} \ldots a_1a_0$, where $n \in \mathbb N$ and $m = a_k + a_{k -1} + \ldots + a_1 + a_0$ , then $n \equiv m \pmod 3$. My strategy is to prove $3|m$ first. Then show $3|m \to 3|n$. $3|n - m$ would then just follow. – Erbolat Jul 26 '14 at 16:39
  • @Erbolat Ok, that's what I surmised. I've explained how to prove this (casting out threes) in an answer to your subsequent question. – Bill Dubuque Jul 26 '14 at 21:06

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We need to know more about the $a_i$ but you can simplify the calculations. Let:

$$S_0 = \lbrace a_i\ \vert \ a_i \equiv 0 \pmod{3}\rbrace$$ $$S_1 = \lbrace a_i\ \vert \ a_i \equiv 1 \pmod{3}\rbrace$$ $$S_2 = \lbrace a_i\ \vert \ a_i \equiv 2 \pmod{3}\rbrace$$

Then $3\vert m$ iff $3$ divides $\vert S_1\vert + \vert S_2\vert$

Darth Geek
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