2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Very clean, totally escaped me! $\endgroup$ – André Armatowski Apr 6 at 16:20
2
$\begingroup$

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 $11$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.