Divisibility rules/Rule for 11 proof

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

Proof

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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Divisibility rules/Rule for 11 proof

Proof

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