$11$ 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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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.

• 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. is a poor mathematician 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

$$\rm(x+1)\ (a_n\ x^n +\:\cdots\:+a_1\ x + a_0)\ =\ a_n\ x^{n+1}+ (a_n+a_{n-1})\ x^{n}+\:\cdots\:(a_1+a_0)\ x+ a_0$$

• Thank you sir. I need little bit more explanation. Could you? – paul Oct 11 '11 at 6:01