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Let $\mathcal{N} \ $ be a natural number of the form $\mathcal{N}=\textrm{dcba}$ ($a$ being the number of units $b$ the tens digit $c$ the hundreds digit and $d$ the thousands digit).

On what condition $\mathcal{N}$ it is divisible by $11$ ?

my attempt:

note that
A number is divisible by 11 :

If you sum every second digit and then subtract all other digits and the answer is:

$0$, or divisible by $11$

By Mr Bill Dubuque i'll try to answer it

indeed,

Our Goal is to show that : $$ P: 11/ (d-c)+(b-a) $$

$\textrm{P} \Longleftrightarrow(d-c)+(b-a) \equiv 11 (\textrm{mod } ) $ or we have for $\ {\rm mod}\ 11\!:\ 10\equiv -1\,\Rightarrow\,10^2\equiv 1\,\Rightarrow\,10^3\equiv -1,\ $ by the $ $ Congruence Power Rule,

therefore $\ a + 10\,b + 10^2 c + 10^3 d \,\equiv\, a-b+c -d,\ $ by the $ $ Congruence Sum, Product Rules.

Or $\,\ P(10)\equiv P(-1)\ $ for $\ P(x) = a + bx + cx^2\! + dx^3,\ $ by the $ $ Polynomial Congruence Rule.

I'm still trying to complet the proof

Is there many ways to prove this and which one is the best and easy to see it ?

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Educ
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    How did you get $(d-c)+(b-a) = 909$? Since $a,b,c,d$ are digits between $0$ and $9$, it is impossible for $(d-c)+(b-a)$ to be larger than $18$. – JimmyK4542 Sep 01 '14 at 22:23

3 Answers3

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Note that $N = 1000d+100c+10b+a$ $= (1001d+99c+11b)+(-d+c-b+a)$ $= 11(91d+9c+b)-[(d-c)+(b-a)]$.

Hence, $N$ is a multiple of $11$ iff $(d-c)+(b-a)$ is also a multiple of $11$.

JimmyK4542
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Hint $\ {\rm mod}\ 11\!:\ 10\equiv -1\,\Rightarrow\,10^2\equiv 1\,\Rightarrow\,10^3\equiv -1,\ $ by the $ $ Congruence Power Rule,

therefore $\ a + 10\,b + 10^2 c + 10^3 d \,\equiv\, a-b+c -d,\ $ by the $ $ Congruence Sum, Product Rules.

Or $\,\ P(10)\equiv P(-1)\ $ for $\ P(x) = a + bx + cx^2\! + dx^3,\ $ by the $ $ Polynomial Congruence Rule.

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Bill Dubuque
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  • could you please be more clear ? make it more details – Educ Sep 01 '14 at 22:52
  • @Educ What is not clear to you? – Bill Dubuque Sep 01 '14 at 22:56
  • In general for me i begin with the goal of proof what i want to prove after that i go step by step . Thanks a lot – Educ Sep 01 '14 at 22:59
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    @Educ $\ {\rm mod}\ 11!:\ n = \sum n_k 10^k = \color{#c00}{P(10)\equiv P(-1)} = \sum n_k (-1)^k.,$ Therefore, in particular, $\tag{}$ $, n = \color{#c00}{P(10)}\equiv 0\iff \color{#c00}{P(-1)}\equiv 0,\ $ i.e. $\ 11\mid n \iff 11\mid P(-1) = n_0 - n_1 + n_2 - n_3 +,\cdots $ $\tag{}$ – Bill Dubuque Sep 01 '14 at 23:14
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    @Educ Please tell me precisely what is not clear to you. Are you already familiar with congruences? – Bill Dubuque Sep 02 '14 at 16:21
  • yes i do but the problem i understand by details step by step not by hint sorry to distrub you – Educ Sep 02 '14 at 16:22
  • @Educ Do you understand the proof in the above comment? That's the proof that I hinted (except the comment does it for any number of digits, i.e. any degree polynomial $,P(x)).\ \ $ – Bill Dubuque Sep 02 '14 at 16:28
  • @Educ I was in the process of answering in your latest question, but you deleted it. You should avoid doing that, since it wastes the time of those composing answers. – Bill Dubuque Sep 02 '14 at 16:44
  • Now i see your proof Thanks a lot – Educ Sep 02 '14 at 17:40
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Proof by induction.

$0$ satisfies the rule.

Create multiples of 11 by successively adding 11.

In the simplest case this adds 1 to both b and a, leaving $S = d -c + b - a$ unchanged.

Any carrying operations add 1 to a digit and subtract 10 from a digit to its right. These operations also preserve $S \, (\mod \, 11)$.

MartinG
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