<s>[INST] Why do some divisibility rules work only in base 10? Aside from the zero rule (a number ending with zero means it is divisible by the base number and its factors), why is it that other rules cannot apply in different bases?
In particular for 3 one can just sum all digits and see if it is divisible. What relation exists between 10 and 3 to have this property? Does this  property exist in another base?
Also:
Are there divisibility rules for base 2?
Are there general divisibility formulae for all bases?
 [/INST] There are other rules for other bases.
To understand why the rule for $3$ works in base $10$, you need to write your number $x=d_n\times 10^n+\dots+d_1\times 10^1+d_0$ where the $d_i$ are the digits of $x$ in base $10$.
You can notice that $10=3\times 3+1$, $10^2=3\times 33+1$, $10^n=3...3\times 3+1$ (where $3...3$ has $n-1$ digits equal to $3$).
So $$\begin{align}x &=d_n\times (3...3\times 3+1)+\dots+d_1\times (3\times 3+1)+d_0 \\ &= (d_n\times 3...3 + \dots+d_1\times 3)\times 3 +d_n+\dots+d_1+d_0\end{align}$$
As you can see, the first term is divisible by $3$, so $x$ is divisible by $3$ if and only if $d_n+\dots+d_1+d_0$ is divisible by $3$.
So the reason that this rule works is the fact that the rest of the division of a power of $10$ by $3$ is $1$.
In base $5$, the same rule applies with $2$ (because the powers of $5$ are odd). For example, in base $5$, $12$ is not divisible by $2$ ($1+2=3$ is not divisible by $2$) but $13$ and $431$ are.
To find rules like this by yourself for a base $b$ and a divisor $d$, you need to study how the powers of $b$ behave when divided by $d$.
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