We know that a number is divisible by $11$ if the difference of sum of odd placed digits and the sum of even placed digits is divisible by $11$.
For example, the number $1023$ is divisible by $11$:
$1+2=3$
$0+3=3$
$3-3=0$
$0$ is divisible by $11$. so $1023$ is divisible by $11$.
Also we can say $1-0+2-3=0$.
On the other hand there are unique rules for divisibility by odd prime numbers. For example, to find if a number is divisible by $7$, we should multiply the unit digit by $2$, then substract the remaining number from that. For example, for $343$ we have $34-6=28$ and $28$ is divisible by $7$.
My question: Is there such a rule for divisibility by $11$ as well?
There is one for each prime number different from 2 and 5. Suppose $10a + b$ is your number, where $b$ is just the last digit, and $p$ is your prime. You have that $$ p|10a + b \iff p| a + (10)^{-1}b $$ In other words, you just have to find the inverse of $10$ (i.e. a number $-p<k<p$ such that $10 k$ is a multiple of $p$ plus 1) modulo $p$.
For $p=7$ we have $k=5$ since $5*10=50 = 49+1$, but also $k=-2$ since $-2*10 = -20 = -21 +1$.
For $p=11$ we have $k=-1$ since $-1*10 = -10 = -11 +1$. In other words, you just need to subtract the unit digit from the rest of the number.