Divisibility by 11

Theorem
Let $N \in \N$ be expressed as:


 * $N = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$

Then $N$ is divisible by $11$ $\displaystyle \sum_{r \mathop = 0}^n \paren {-1}^r a_r$ is  divisible by $11$.

That is, a divisibility test for $11$ is achieved by alternately adding and subtracting the digits and taking the result modulo $11$.

Proof
As:
 * $10 \equiv -1 \pmod {11}$

we have:
 * $10^r \equiv \paren {-1}^r \pmod {11}$

from Congruence of Powers.

Thus:
 * $N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {11}$

from the definition of Modulo Addition.

The result follows.