Trick to find multiples mentally We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.
A few months ago, I heard a simple and elegant way to find multiples of $7$:
Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.
Example:
$21553$
Cut digits into pairs:
$2, 15, 53$
Multiply $53$ by $1, 15$ by $2, 2$ by $4$:
$8, 30, 53$
Add:
$8+30+53=91$
As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.
This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$
I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.
Exemple with $91$:
$91$
$9, 1$
$9\cdot3, 1\cdot1$
$27, 1$
$28$
My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?

I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$ z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$
Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have, 
$$ 10 \equiv 3, 10^2 \equiv 2, 10^3 \equiv -1, 10^4 \equiv -3, 10^5 \equiv -2, 10^6 \equiv 1,$$
The successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$ r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\cdots$$ is divisible by $7$
To understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \times (10-3) + a_2 \times (10^2-2) + a_3 \times (10^3+1) + \cdots + a_6 \times (10^6-1)
 + \cdots $$

Since all this numbers $ (10-3),(10^2-2),(10^3+1),\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.
Using this approach we can derive divisibility of any integer.
