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I'm trying to teach myself some number theory. In my textbook, this proof is given:

Example (2.3.1) Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3.

Let $n=a_0a_1\ldots a_k$ be the decimal representation of an integer $n$. Thus $n=a_k+a_{k-1}10+a_{k-2}10^2+\cdots+a_010^k$ where $0\le a_i<10$ The key observation is that $10\equiv1\pmod3$, i.e., $[10]=[1]$. Hence $[10^i]=[10]^i=[1]$, i.e., $10^i\equiv1\pmod 3$. The assertion is an immediate consequence of this congruence.

I don't understand the last statement. Why does it follow from that congruence?

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John L.
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  • Do you understand that $10^i \bmod 3 = 1$? – user251257 Sep 30 '15 at 03:23
  • Yes. I'm a little weary when it comes to congruences in general, but I understand that 10^i = 1 (mod 3) -- with the equal sign here denoting equivalence. Hence a(10^i) = a([1]) = a. Is this correct? – John L. Sep 30 '15 at 03:26
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    yes ... (just the notation is wonky, better: $a \cdot 10^i \equiv a \cdot 1 \equiv a \pmod 3$.) – user251257 Sep 30 '15 at 03:28

3 Answers3

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A simple way to see this (that actually generalizes nicely to Fermat's little theorem):

$$10 - 1 = 9 = 9 \times 1$$ $$100 - 1 = 99 = 9 \times 11$$ $$1000 - 1 = 999 = 9 \times 111$$ In general $$10^n - 1 = 9 \times \underbrace{111...111}_\mbox{$n$ times}.$$ This is just the algebraic identity $$x^n - 1 = (x - 1)(x^{n-1} + x^{n - 2} + ... + x + 1)$$ when $x = 10$. The identity is easy to prove - just multiply it out term by term. All but the first and last terms cancel. Thus any power of $10$ less $1$ is divisible by $9$, and therefore also by $3$.

Now consider a multi-digit natural number, $43617$ for example. $$\begin{align} 43617 &= 4\times 10^4 + 3 \times 10^3 + 6 \times 10^2 + 1 \times 10 + 7 \\ &= 4\times 9999 + 3 \times 999 + 6 \times 99 + 1 \times 9 + (4 + 3 + 6 + 1 + 7) \end{align}$$

Every term on the right other than the sum of the digits is divisible by $3$. So the remainder when dividing the original number by $3$ and the sum of the digits by $3$ must be the same.

Alexander Belopolsky
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Paul Sinclair
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    The OP question wasn't about proving the statement. It was to explain the last line of the proof given. – fleablood Sep 30 '15 at 05:37
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    regardless, thank you so much for having taken the time to write this out! that helps me understand it from a different angle, and I really appreciate that! :) – John L. Sep 30 '15 at 11:36
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    @fleablood - I was aware of that, and you had answered the question adequately already. But I thought that the different approach might help him improve his understanding of what is going on. Sometimes we get so caught up in manipulating symbols that it is easy to lose sight of what they mean. – Paul Sinclair Sep 30 '15 at 16:57
  • Paul Sinclair, indeed! In the terse proof of the OP, the $10^i = 1 mod (3)$ leading immediately to n = sum of digits mod (3) took me more than a few moments to parse. (Because each digit times $10^ii$ is eq. the digit mod 3! Duh!) and the phrasing of the last sentence ... well, it didn't stump me as long but it wasn't as obvious as a spoken language would have been. But both these proofs are slicker and more elegant the pre-algebra proof I tend to give. – fleablood Sep 30 '15 at 17:50
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n = $a_0$ + .... + $a_n$ mod (3) means that n and the sum of the digits will be equivalent to the same number modulo 3. If this number is 0 then n and the sum of the digits will both be divisible by 3. If the number isn't 0 (or any other multiple of 3) neither n nor the sum of the digits will be divisible by 3.

fleablood
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The assertion is an immediate consequence of this congruence.

Since any power of $10$ is congruent to $1$ (mod $3$),

$$n=a_k+a_{k-1}10+a_{k-2}10^2+\cdots+a_010^k\equiv a_k+a_{k-1}+a_{k-2}+\cdots+a_0\pmod3.$$

In other words, the residual of dividing $n$ by $3$ is the same as the residual of dividing the sum of its digits by $3$. In the case of zero residual, we get the sought assertion: $n$ is divisible by $3$ iff the sum of its digits is divisible by $3$.

Alexander Belopolsky
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