11 divisibility

We know that 1331 is divisible by 11. As per the 11 divisibility test, we can say 1331 is divisible by 11. However we cannot get any quotient. If we subtract each unit digit in the following way, we can see the quotient when 1331 is divided by 11.

1331 $\implies$ 133 -1 = 132 132 $\implies$ 13 - 2 = 11 11 $\implies$ 1-1 = 0, which is divisible by 11. Also the quotient is, arrange, all the subtracted unit digits (in bold and italic) from bottom to top, we get 121. Which is quotient. I want to know how this method is working? Please write a proof.

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Hello! I see that you are new to math.stackexchange.com. A few tips when posting questions: When using verbs like "to want", please use the conditional form of the verb (Say I would like... and not I want) or be polite and say "Can I" or "May I know"... Also, you may consider rephrasing your question above as it is not clear exactly what you are asking about. Cheers! –  user38268 Oct 11 '11 at 5:50
@Benjamin, I want to know what is wrong with saying "I want" as it doesn't seem impolite at all. Also the question seems clear enough. –  Dan Brumleve Oct 11 '11 at 6:05

Look how the method works for the number $abcde$. You substract $e$ from $abcd$, getting $a'b'c'd'$, and you apply the method to $a'b'c'd'$. If the method works for $a'b'c'd'$, it yields $n'$ such that $a'b'c'd'=11\times n'$. But $a'b'c'd'0$ is $a'b'c'd'\times 10=11\times 10\times n'$ and $abcde=a'b'c'd'\times 10+e\times 11$ hence $abcde=11\times 10\times n'+11\times e$ is $11\times(10\times n'+e)$. This proves that $n=10\times n'+e$ is indeed the correct answer for $abcde$.

Edit (Upon OP's request, the same proof, with more apparatus but with zero more mathematics.)

Look how the method works for the number $N=a_ka_{k-1}\cdots a_2a_1a_0$ with $k\geqslant1$. You substract $a_0$ from $a_ka_{k-1}\cdots a_2a_1$, getting $M=b_kb_{k-1}\cdots b_2b_1$, and you apply the method to $M$. If the method works for $M$, it yields $m$ such that $M=11\times m$. But $b_kb_{k-1}\cdots b_2b_10$ is $M\times 10=11\times 10\times m$ and $N=M\times 10+a_0\times 11$ hence $N=11\times 10\times m+11\times a_0$ is $11\times(10\times m+a_0)$. This proves that $n=10\times m+a_0$ is indeed the correct answer for $N$ if $m$ was the correct answer for $M$. A recursion on the number of digits of $N$ yields the result.

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I want by generalization. Do not take an example consisting for 5 digits or limited digits. Proof should be in general –  paul Oct 11 '11 at 6:02
See edit. (But you really have to stop giving orders to people, you know?) –  Did Oct 11 '11 at 6:12
@paul, did you just downvote my answer? –  Did Oct 11 '11 at 6:13
@Didier: he couldn't have; he doesn't have sufficient reputation to downvote answers. It is quite annoying that the downvoter couldn't be bothered to explain his downvote. Bleh. –  Ｊ. Ｍ. Oct 11 '11 at 6:18
@J.M. Thanks for reminding me this feature of the site (I knew this once but seem to have forgotten it since). By the way, I agree with the penultimate sentence of your comment. Hmmm... and in fact, with the last one too. :-) –  Did Oct 11 '11 at 6:30
HINT $\$ Specialize $\rm\ x = 10\$ below
$$\rm(x+1)\ (a_n\ x^n +\:\cdots\:+a_1\ x + a_0)\ =\ a_n\ x^{n+1}+ (a_n+a_{n-1})\ x^{n}+\:\cdots\:(a_1+a_0)\ x+ a_0$$