Can divisibility rules for digits be generalized to sum of digits Suppose that we are given a two digit number $AB$, where $A$ and $B$ represents the digits, i.e 21 would be A=2 , B=1. I wish to prove that the sum of $AB$ and $BA$ is always divisible by $11$. 
My initial idée was to use the fact that if a number is divisible by $11$ then the sum of its digits with alternating sign is also divisible by 11. For example
$$1-2+3-3+2-1=0  $$
so $11$ divides $123321$. So my proof would then be to consider the two digit number $(A+B)(B+A)$ or $CC$ which clearly is divisible by $11$ by the above statement if $C$ is $1$ through $9$. However, I am having truble justifying the case were $A+B$ is greater than or equal to $10$ and it got me wondering if the more generel is true: Let $ABCD...$ be a $n-digit$ number, if $$A-B+C-... \equiv 0 (mod \ 11)$$
then  $$S=\sum_{k=1}^{n}(A+B+C+...)10^{k} \equiv0(mod \ 11)$$ 
I am not really familliar with the whole congruence thingy, so incase the above is trivial I would be greatful on some source which could aid the solving of the above . Any tips or suggestion are also very welcome
 A: You can easily push through what you were trying though. Say your numbers are $AB$ and $BA$. If $A+B\gt 10$, then write it as $A+B=10+c$; note that $0\leq c\leq 8$, because two digits cannot add to $19$. That means that when you do the carry, the second digit is $(c+1)$, and so $AB+BA$ will be a three digit number: $1$, then $c+1$, and then $c$. At this point, your test gives you $1-(c+1)+c = 0$, so you get a multiple of $11$.
A: More generally, recall that the radix $\rm\,b\,$ digit string $\rm\ d_n \cdots\ d_1\ d_0\ $ denotes a polynomial expression $\rm\ P(b) = d_n\ b^n +\:\cdots\: + d_1\ b + d_0,\, $ where $\rm\ P(x) = d_n\ x^n +\cdots+ d_1\ x + d_0.\, $ Recall the reversed (digits) polynomial is $\rm\ {\bf \tilde {\rm P}}(x) = x^n\ P(1/x).\,$ If $\rm\:n\:$ is odd  the Polynomial Congruence Rule yields $$\rm\: mod\ \ b\!+\!1:\ \ \color{#c00}{b\equiv -1}\ \Rightarrow\ {\bf \tilde {\rm P}}(b) = \color{#c00}b^n\ P(1/\color{#c00}b) \equiv (\color{#c00}{-1})^n P(\color{#c00}{-1})\equiv {-}P(-1),\:$$ therefore we conclude that $\rm\ P(b) + {\bf \tilde {\rm P}}(b)\equiv P(-1)-P(-1)\equiv 0.\,$ OP is case $\rm\,b=10,\ n=1$.
Remark $ $ Essentially we have twice applied the radix $\rm\,b\,$ analog of casting out elevens (the  analog of casting out nines).
A: It's simpler than you are making it...and no congruences are needed:
We have $$\overline {AB}=10A+B \quad \&\quad \overline {BA}=10B+A$$
It follows that $$\overline {AB}+\overline {BA}=11\times (A+B)$$ and we are done.
