# 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

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).

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.

• Very clean, totally escaped me! – André Armatowski Apr 6 at 16:20

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 $$11$$.