Particular number is divisible by 11

Question

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 ?

Answer 1

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 $11$.

Answer 2

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.

Answer 3

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 $S \, (\mod \, 11)$.