How to prove the divisibility rule for $3\, $ [casting out threes]

HINT: Suppose that you have a four-digit number $n$ that is written $abcd$. Then

$$\begin{align*} n&=10^3a+10^2b+10c+d\\ &=(999+1)a+(99+1)b+(9+1)c+d\\ &=(999a+99b+9c)+(a+b+c+d)\\ &=3(333a+33b+3c)+(a+b+c+d)\;, \end{align*}$$

so when you divide $n$ by $3$, you’ll get

$$333a+33b+3c+\frac{a+b+c+d}3\;.$$

The remainder is clearly going to come from the division $\frac{a+b+c+d}3$, since $333a+33b+3c$ is an integer.

Now generalize: make a similar argument for any number of digits, not just four. (If you know about congruences and modular arithmetic, you can do it very compactly.)

$\begin{eqnarray} \rm{\bf Hint}\ \ &&\rm3\ \ divides\ \ a\! +\! 10\,b\! +\! 100\, c\! +\! 1000\,d\! + \cdots\\ \iff &&\rm 3\ \ divides\ \ a\! +\! b\! +\! c\! +\! d\! +\! \cdots +\color{#c00}9\,b\! +\! \color{#c00}{99}\,c\! +\! \color{#c00}{999}\,d\! + \cdots\\ \iff &&\rm3\ \ divides\ \ a\! +\! b\! +\! c\! +\! d + \cdots\ \ by\ \ 3\ \ divides\ \ \color{#c00}{9,\ 99,\ 999,\,\ldots}\end{eqnarray}$

Above we used that $\rm\ n + 3m\ $ is divisible by $\rm\,3\iff n\:$ is divisible by $\,3.$

$1$. First prove that $3 \mid 10^n - 1$ (By noting that, $10^n - 1 = (10-1)(10^{n-1} + 10^{n-2} + \cdots + 1)$).

$2$. Now any number can be written in decimal expansion as $$a = a_n 10^n + a_{n-1} 10^{n-1} + \cdots + a_1 10^1 + a_0$$

$3$. Note that $a_k 10^k = a_k + a_k (10^k-1)$. Hence, $$a = \overbrace{(a_n + a_{n-1} + \cdots + a_0)}^{b} + \underbrace{\left(a_n (10^n-1) + a_{n-1} (10^{n-1}-1) + \cdots + a_1 (10^1-1) \right)}_{c}$$

$4$. We have $a=b+c$ and $3 \mid c$. Now conclude that $3 \mid a \iff 3 \mid b$.