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 $\displaystyle \sum_{r=0}^n \left({-1}\right)^r a_r$ 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:

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

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.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): Exercise $2.9$