Divisibility by 7 What is the fastest known way for testing divisibility by 7? Of course I can write the decimal expansion of a number and calculate it modulo 7, but that doesn't give a nice pattern to memorize because 3 is a primitive root. I'm looking for alternative ways that can help you decide when a number is divisible by 7 by hand.
I'm sorry if this is a duplicate question, but I didn't find anything similar on the site.
 A: Another way to do this is to use a divisibility graph (see: How does the divisibility graphs work?). They're not to difficult to remember/generate for small numbers.

To compute mod 7 for n: Start at 0. For each digit x in n traverse x black arrows in the graph then, in between digits, follow the blue arrows.
example: 
Take 6594:


*

*Digit is 6, traverse 6 black arrows, ending on node 6 then follow blue arrow to node 4

*Digit is 5, traverse 5 black arrows, ending on node 2 then follow blue arrow to node 6

*Digit is 9, traverse 9 (or 2 as 9 mod 7 = 2) black arrows, ending on node 1 then follow blue arrow to node 3

*Digit is 4, traverse 4 black arrows, ending on node 0


6594 mod 7 = 0
A: One rule I use pretty much is: 
If you double the last digit and subtract it from the rest of the number and the answer is:
0, or divisible by 7 then the number itself is divisible by 7.
Example: 


*

*672 (Double 2 is 4, 67-4=63, and 63÷7=9) Yes

*905 (Double 5 is 10, 90-10=80, and 80÷7=11 3/7) No


If the number is too big you can repeat until you find the solution. 
A: Here are some pointers:
Divisibility by 7
Divisibility rules
A: I created this algorithm that is very quick:
N = a,bcd;
a' = (- cd mod 7 + a) mod 7;
If 7|a'b then 7|N;
If N is larger repeat the procedure.
It works because - cd mod 7 ≡ 6 cd; 6 cd goes to the place value of the thousands. Then N is submitted to the addition of 6,000 cd and to the subtraction of 1 cd: 6,000 - 1 = 5,999 and 7|5,999.
Each time the procedure is applied a multiple of 7 is added to N.
With practice it may be applied entirely through mental calculation.
Watch this video that shows the application of the algorithm:
https://www.youtube.com/watch?v=d8Bmf9BNonU
A: The best way to test if a number is divisible by any other number is by deducting the number n times and check whether the remainder is divisible by n. for example 861-7n/7. Here 7n must be lesser or equal to 861. 
A: The rule $7 \mid 10n+d \iff 7 \mid n-2d$ works fine when the number $10n+d$ is "small" but it would be too much work for, say, the number $2726394118$
However, the fact that $7 \mid 1001$ can be used to speed things up a bit.
First, we note that $1000^m \equiv (-1)^m \pmod{1001}$. This implies the following method.


*

*Break the number up into periods of $3$ digits (from right to left).


$$2726394118 \mapsto 2 \quad 726 \quad 394 \quad 118$$


*

*Sum the odd periods


$$118 + 726 = 844$$


*

*Sum the even periods


$$2 + 394 = 396$$


*

*Compute the difference


$$844-396 = 448$$


*

*The original number is divisible by $7$ if and only if the difference divisible by $7$.


$$448 \mapsto 44-16 = 28$$
Since $7$ divides $28$, then $7$ divides $2726394118$.
