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 $N = a_ka_{k-1}\cdots a_1a_0$ where the $a_i$ are base-ten numbers. Then $N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0.$

Note that $10\equiv -1\pmod{11}$ . Thus $10^k a_k\!\! +\!\!10^{k-1} a_{k-1}\! +\! \cdots + 10a_1 + a_0 \equiv (-1)^k a_k + (-1)^{k-1} a_{k-1} + \cdots -a_1 + a_0 \pmod {11}.$

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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