Divisibility Test for 7, 11 and 13

Theorem
Mark off the integer $N$ being tested into groups of $3$ digits.

Because of the standard way of presenting integers, this may already be done, for example:
 * $N = 22 \, 846 \, 293 \, 462 \, 733 \, 356$

Number the groups of $3$ from the right:
 * $N = \underbrace{22}_6 \, \underbrace{846}_5 \, \underbrace{293}_4 \, \underbrace{462}_3 \, \underbrace{733}_2 \, \underbrace{356}_1$

Considering each group a $3$-digit integer, add the even numbered groups together, and subtract the odd numbered groups:


 * $22 - 846 + 293 - 462 + 733 - 356 = -616$

where the sign is irrelevant.

If the result is divisible by $7$, $11$ or $13$, then so is $N$.

In this case:
 * $616 = 2^3 \times 7 \times 11$

and so $N$ is divisible by $7$ and $11$ but not $13$.

Proof
Let $N$ be expressed as:
 * $N = \displaystyle \sum_{k \mathop = 0}^n a_k 1000^k = a_0 + a_1 1000 + a_2 1000^2 + \cdots + a_n 1000^n$

where $n$ is the number of groups of $3$ digits.

We have that:
 * $1000 \equiv -1 \pmod {1001}$

Hence from Congruence of Powers:
 * $1000^r \equiv \left({-1}\right)^r \pmod {1001}$

Thus:
 * $N \equiv a_0 + \left({-1}\right) a_1 + \left({-1}\right)^2 a_2 + \cdots + \left({-1}\right)^n a_n \pmod {1001}$

from the definition of Modulo Addition.

Then we note that:
 * $1001 = 7 \times 11 \times 13$

and the result follows.