Divisibility rules and congruences Sorry if the question is old but I wasn't able to figure out the answer yet. 
I know that there are a lot of divisibility rules, ie: sum of digits, alternate plus and minus digits, etc... but how can someone derive those rules for any number $n$ let's say. I know it could be done using congruences, but how ?
Thank you !
 A: A positive integer written as  $x = d_k \ldots d_1 d_0$ (in base 10) is really $\sum_{j=0}^k d_j 10^j$.  Suppose $10^j \equiv m_j \mod n$.  Then $x$ is divisible by $n$ if and only if $\sum_{j=0}^k d_j m_j$ is divisible by $n$.  Assuming $n$ and 10 are coprime, $10^j$ is periodic mod $n$, the minimal period being a divisor of $\varphi(n)$. 
For example, in the case $n=7$, we have $m_0 = 1$, $m_1 = 3$, $m_2 = 2$, $m_3 = -1$, $m_4 = -3$, $m_5 = -2$, and then it repeats.  So $x$ is divisible by 7 if and only if $(d_0 -  d_3 + d_6 - d_9 + \ldots) + 3 (d_1 - d_4 + d_7 - d_{10} + \ldots) + 2 (d_2 - d_5 + d_8 - d_{11} + \ldots)$ is.   
A: Here's one example... maybe it will help you to show that something is divisible by 3 iff its digits are divisible by 3. Write your number in expanded notation[using Robert Israel's notation]:
$N=\displaystyle\sum_{j=0}^n d_j10^j, d_j\in\{0,1,\dotsc,9\}$ (assume $d_n\neq 0$)
We want to know when this number is divisible by 3; said equivalently, when this number is congruent to $0\pmod{3}$. I claim it's when the sum of the digits is divisible by 3. To show this, take our number $N\pmod{3}$:
$\displaystyle\sum_{j=0}^n d_j10^j\equiv \displaystyle\sum_{j=0}^n d_j1^j =\displaystyle\sum_{j=0}^n d_j\pmod{3}\text{ and we see that}$
When $\displaystyle\sum_{j=0}^n d_j\equiv0\pmod{3},~\displaystyle\sum_{j=0}^nd_j10^j$ is divisible by 3.
A: We need not memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \ $ is $\rm\: (\color{#0a0}{d_2\cdot\color{#90f}{10} + d_1})\ 10 + d_0\ \equiv\ (\color{#0a0}{d_2\cdot 3 + d_1})\ 3 + d_0\ (mod\ 7)\ $ since $\rm\ \color{#90f}{10}\equiv\color{#0a0}3\ (mod\ 7).\:$ Hence we compute the remainder $\rm\ (mod\ 7)\ $ as follows. Start with the leading digit then repeatedly apply the operation: $\rm\color{#0a0}{multiply\ by\ 3\ then\ add\ the\  next\ digit}$, doing all of the arithmetic $\!\bmod 7$.
For example, let's use this algorithm to compute $\, 43211\bmod 7.\,$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ \color{#0a0}{d_n\ d_{n-1}}\ $ by $\rm\, \color{#0a0}{(\color{#000}3\, d_n + d_{n-1})}\bmod 7,\,$ namely
$$\begin{array}{rrl}\bmod 7\!:\  &\color{#0A0}{4\ 3}\ 2\ 1\ 1^{\phantom{|^{|}}}\!\!\!&\\
\equiv\!\!\!\! &\color{#c00}{1\ 2}\ 1\ 1 &\!{\rm by}\ \ \:\! \smash[t]{\overbrace{3\cdot \color{#0a0}4 + 
 \color{#0a0}3}^{\rm\textstyle\color{#0a0}{\,\color{#000} 3\,\ d_n\! + d_{n-1}}\!\!\!\!\!\!\!}} \equiv\ \color{#c00}1\\
\equiv\!\!\!\! &\color{#0af}{5\ 1}\ 1&\!{\rm by}\ \ \ 3\cdot \color{#c00}1 + 
 \color{#c00}2\ \equiv\ \color{#0af}5\\
\equiv\!\!\!\! & \color{#f60}{2\ 1}&\!{\rm by}\ \ \ 3\cdot \color{#0af}5 + 
 \color{#0af}1\ \equiv\ \color{#f60}2\\
\equiv\!\!\!\! &\color{#8d0}0&\!{\rm by}\ \ \ 3\cdot \color{#f60}2 + 
 \color{#f60}1\ \equiv\ \color{#8d0}0
\end{array}\qquad\qquad\quad\ \,   $$
Hence $\rm\ 43211\equiv 0\pmod{\!7},\,$ indeed $\rm\ 43211 = 7\cdot 6173.\:$ Generally the modular arithmetic is simpler if we use least magnitude residues, e.g. $\rm\, \pm\{0,1,2,3\}\ \:(mod\ 7),\,$ by allowing negative digits, e.g. here. Note that for modulus $11$ or $9\:$ the above  method reduces  to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9$).
A: Bill Dubuque's answer gave me the motivation to calculate the $\text{modulus-}7$ residue 'on a computer'.
In $\text{base-}2$
$\quad 43211 =1010100011001011_2$
Computing the $\text{mod-}7$ residue of $43211$:
Initialize the algorithm input, $1010100011001011 \sim [1]010100011001011$.
$\quad [1]010100011001011; \; [2]10100011001011; \; [5]0100011001011; \; [3]100011001011;$
$\quad [0]00011001011; [0]11001011; \; [1]1001011; \; [3]001011; \; [6]01011; \; [5]1011;$
$\quad [4]011; \; [1]11; \; [3]1; \; [0]$
So $43211 \equiv 0 \pmod7$.
