Difference between revisions of "Divisibility rules/Rule for 11 proof"

Revision as of 08:18, 16 August 2006

A number $N$ is divisible by 11 if the alternating sum of the digits is divisible by 11.

An understanding of basic modular arithmetic is necessary for this proof.

Let $N = a_na_{n-1}\cdots a_1a_0$ where the $a_i$ are base-ten numbers. Then $N = 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10 a_1 + a_0.$

Note that $10\equiv -1\pmod{11}$ . Thus

$10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10a_1 + a_0 \equiv (-1)^n a_n + (-1)^{n-1} a_{n-1} + \cdots -a_1 + a_0.$

This is the alternating sum of the digits of $N$ , which is what we wanted.

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 Let <math>N = a_na_{n-1}\cdots a_1a_0</math> where the <math>a_i</math> are [[base numbers | base-ten]] numbers.  Then <math>N = 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10 a_1 + a_0.</math>
-Note that <math>10\equiv -1\pmod{11}</math>.  Thus <center><math> 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10a_1 + a_0 \equiv (-1)^n a_n + (-1)^{n-1} a_{n-1} + \cdots -a_1 + a_0. </math><center>
+Note that <math>10\equiv -1\pmod{11}</math>.  Thus <center><math> 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10a_1 + a_0 \equiv (-1)^n a_n + (-1)^{n-1} a_{n-1} + \cdots -a_1 + a_0. </math></center>
 This is the alternating sum of the digits of <math>N</math>, which is what we wanted.