Divisibility Test for 7, 11 and 13

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

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

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

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 the number you started with.

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

and so the original number is divisible by $7$ and $11$ but not $13$.