Divisibility Rules for Bases other than $10$ I still remember the feeling, when I learned that a number is divisible by $3$, if the digit sum is divisible by $3$. The general way to get these rules for the regular decimal system is asked/answered here: Divisibility rules and congruences. Now I wonder, what divisibility rules an alien with $12$ (or $42$) fingers would come up with?
So let $n=\sum_k c_k b^k$ be the representation with base $b$. Looking at some examples shows indicate that $n$ is divisible by $b-1$, if $\sum_k c_k$ is. This seems to be a poor man's extension to the decimal divisibility by $9$ rule.
The answer to the above mentioned question, says that "One needn't memorize motley exotic divisibility tests. ". Motley exotic divisibility tests are very welcome here!
 A: Claim 1:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n+1$'. Let '$s$' denote the sum of digits of '$a$' expressed in base '$n+1$'.  Now $n|a \iff n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $14$.
$$611 = 3 \times 14^2 + 1 \times 14^1 + 9 \times 14^0 = (319)_{14}$$
where $(319)_{14}$ denotes that the decimal number $611$ expressed in base $14$. The sum of the digits $s = 3 + 1 + 9 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n+1$'.
$$a = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n+1 & \equiv 1 \pmod n\\
(n+1)^k & \equiv 1 \pmod n \\
a_k \times (n+1)^k & \equiv a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0 \\
& \equiv (a_m + a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Claim 2:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n-1$'. Let '$s$' denote the alternating sum of digits of '$a$' expressed in base '$n-1$' i.e. if $a = (a_ma_{m-1} \ldots a_0)_{n-1}$, $s = a_0 - a_1 + a_2 - \cdots + (-1)^{m-1}a_{m-1} + (-1)^m a_m$. Now $n|a$ if and only $n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $12$.
$$611 = 4 \times 12^2 + 2 \times 12^1 + B \times 12^0 = (42B)_{12}$$
where $(42B)_{14}$ denotes that the decimal number $611$ expressed in base $12$, $A$ stands for the tenth digit and $B$ stands for the eleventh digit.
The alternating sum of the digits $s = B_{12} - 2 + 4 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out just like the one above. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n-1$'.
$$a = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n-1 & \equiv (-1) \pmod n\\
(n-1)^k & \equiv (-1)^k \pmod n \\
a_k \times (n-1)^k & \equiv (-1)^k a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0 \\
 & \equiv ((-1)^m a_m + (-1)^{m-1} a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Pros and Cons:
The one obvious advantage of the above divisibility rules is that it is a generalized divisibility rule that can be applied for any '$n$'.
However, the major disadvantage in these divisibility rules is that if a number is given in decimal system we need to first express the number in a different base. Expressing it in base $n-1$ or $n+1$ may turn out to be more expensive. (We might as well try direct division by $n$ instead of this procedure!).
However, if the number given is already expressed in base $n+1$ or $n-1$, then checking for divisibility becomes a trivial issue.
A: The test for base-10 divisibility by 11 has a straightforward analogue in other bases. For example, in base 12, 756899 is divisible by 13 because 7+6+9 =  5+8+9. One can extend this to a case that doesn't arise in base 10 because 10+1 is prime: If $n$ is any factor of $b+1$, then one can test for divisibility by $n$ in base $b$ by forming the two alternate-digit sums and checking if they differ by a multiple of $n$. For example consider base 8. Is the number ${7166}_8$ divisible by 3? It is if and only if 7+6 and 1+6 differ by a multiple of 3; obviously, they do, so the answer is yes.
The tests for divisibility by 2 and 5 have analogues in non-prime bases. For example, in base 12, numbers are divisible by 6 if and only if they end in 0 or 6; divisible by 4 if and only if they end in 0, 4, or 8; by 3 if and only if they end in 0, 3, 6, or 9; and by 2 if and only if they in 0, 2, 4, 6, 8, or A. Similarly the test for divisibility by 10 has an obvious analogue. In general, if $n$ divides $b$, then a base-$b$ number $X$ is divisible by $n$ if and only if its last digit is divisible by $n$.
The test for divisibility by 3 has analogues in bases $n$ where $n-1$ is not prime; if $n-1$ is divisible by $k$, you can check for divisibility by $k$ by adding up the base-$n$ digits and checking if the sum is divisible by $k$.  For example, consider 82A in base 11. The sum of these digits is 20, which is even, and is also a multiple of 5; both 2 and 5 divide 11-1=10, so 82A is a multiple of both 5 and 2. But the sum of the digits in 654 is 15, a multiple of 5 but not of 2, so 654 is a multiple of 5 but not of 2.
A: Here is a divisibility rule for $d$ in base $b$, given $d$ and $b$ are relatively prime.
Let $k$ be any integer such that $kb\equiv 1 \pmod d$. Then we can take the last digit of the number we're testing, multiply it by $k$, and add it to the remaining digits, not including the last digit. Then we can repeat the process with the new number formed. Note that all calculations are done in base $b$.
For example, the to find if a number, say $552839$, is divisible by 7 in base 10, we can set $k=-2$, since $(-2)10=-20\equiv 1 \pmod 7$. Then we get the following sequence:
$$552839 \Rightarrow 55265 \Rightarrow 5516 \Rightarrow 539 \Rightarrow 35$$
Since $35$ is divisible by 7, then the original number $552839$ is also divisible by 7.
