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 $$11$$ iff $$\sum_{r=0}^n \left({-1}\right)^r a_r$$.

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:
 * $$10^r \equiv \left({-1}\right)^r \pmod {11}$$

from Congruence of Powers.

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

from the definition of Modulo Addition.

The result follows.