Divisibility rule of 11

Question

Let $M$ be a natural number with $n+1$ digits; represented by $M=a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}$

Show $M$ is divisible by $11$ if and only if

$$(a_{0}+a_{2}+a_{4}+\ldots)-(a_{1}+a_{3}+a_{5}+\ldots)$$ is also a divisible by 11

here is Solution By Mr Américo Tavares

Let

$$M=a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}=a_{n}10^{n}+a_{n-1}10a^{n-1}+\cdots +a_{2}10^{2}+a_{1}10+a_{0}.$$

For $\mod 11$ we have $10^{0}\equiv 1,10^{1}\equiv 10,10^{2}\equiv 1$, and if $n$ even

$$10^{n}\equiv 1;$$

if $n$ odd $$10^{n}\equiv 10.$$ Thus

$$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )+10(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$$

Since $-10\equiv 1\pmod{11}$, we have $$10\equiv (-(-10))\equiv -1\pmod{11}$$ and

$$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )-(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$$

Could we prove it by the Methods of :

Mr. Mr. JimmyK4542
Mr Bill Dubuque

from here Particular number is divisible by 11

Or Is There another way to prove it ?

@Educ But this is so simple that there is really only one point of view! Anything else is just going to be the same idea presented slightly differently ... — almagest, Sep 02 '14 at 16:54
@almagest i can't be sure for the moment until to see others ways — Educ, Sep 02 '14 at 16:56
@Educ, Related : http://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 — lab bhattacharjee, Sep 02 '14 at 17:56

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

I suspect that the question stems from my use of (radix) polynomials in my linked answer. This does indeed yield a generalization of such divisibility tests. Let's examine the proof closely.

First, notice that the radix representation of an integer is a polynomial function of the radix (base), i.e. if $\,n\,$ has digits $\,n_k \cdots n_1 n_0\,$ in radix $\,b\,$ then this means that

$$n = n_k b^k + \cdots + n_1 b + n_0 = P(b),\ \ P(x) = n_k x^k +\cdots + n_1 x + n_0$$

The test is simply $\ b\!+\!1\mid P(b)\iff b\!+\!1\mid P(-1)\ $ since $\ {\rm mod}\,\ b\!+\!1\!:\ b\equiv -1\,\Rightarrow\, P(b)\equiv P(-1),\ $ by the Polynomial Congruence Rule. Therefore $\, P(b)\equiv 0\iff P(-1)\equiv 0,\ $ as claimed.

The proof works for any polynomial $\,P(x)\,$ with integer coefficients. As such, these tests for divisibility by the radix$\pm1$ (e.g. also casting out nines) may be viewed as special cases of the Polynomial Congruence Rule.

score 1 · Answer 2 · edited Sep 02 '14 at 17:05

$$10=11-1=11k-1 \\100=99+1=9(11)+1=11q+1\\1000=1001-1=11m-1\\10000=9999+1=11n+1\\\text{and so}\\M=a_{n}...a_{3}a_{2}a_{1}a_{0}=a_{0} +10a_{1}+100a_{2}+1000a_{3}+...=\\M=a_{0} +(11-1)a_{1}+(99+1)a_{2}+(1001-1)a_{3}+...=\\M=a_{0} +(-1)a_{1}+(1)a_{2}+(-1)a_{3}+...+11(1a_{1}-9a_{2}+99a_{3}....)=\\M=a_{0} +(-1)a_{1}+(1)a_{2}+(-1)a_{3}+...+11Q\\\text{now divide M by 11}\\M=(a_{0} +(-1)a_{1}+(1)a_{2}+(-1)a_{3}+...)+11Q =M'+11Q\\M/11\rightarrow M' $$

Divisibility rule of 11

2 Answers2