How to know a number is divisible by a given number without using a calculator? My question is simple and comes from my curiousity during studying math.

How to know a number is divisible by $7$ or $13$ without using a calculator? For example, how do we decide intuitively that $11172686$ is divisible (or not) by $7$ or $13$?

Moreover, I am interesting in knowing a simple rules or condition that a number is divisible by a given number without using a calculator? Can you help me answering and explaining this topic? Thanks.
 A: For $7$
$11172686$
Remove last Digit
Now we have $1117268$, Subtract twice of removed digit ($2\times6$) from $1117268$
Repeat this till we get either a known multiple of $7$ or $7$ itself or Zero ($0$)
If you're unable to get any of this, Given number is not divisible by $7$
For $13$
$11172686$ Remove Last digit
Now we have $11172686$, Add 4 times of removed digit to $11172686$ 
Repeat this till we get either a known multiple of $13$ or $13$ itself
If you're unable to get any of this, Given number is not divisible by $13$
A: for the first question you must calculate $686-172+11=525=75\cdot 7$
for the second question we get $686-172+11=525$ and $525$ is not devisible by $13$
A: Specifically for divisions with $7$ and $13$ (and $11$) there are very simple tests available online. 


*

*Division with 7

Take off the last digit and multiply it by 2. 
      Subtract it from the digits you have left. 
      If the answer divides by 7 (or is zero), then your original number will also divide by 7!



*

*Division with 13

Take off the last digit and multiply it by 9. 
      Subtract it from the digits you have left. 
      If the answer divides by 13 (or is zero), then your original number will also divide by 13!

A: There are several "rules", each limited to a number of family of numbers...
For instance, every even number is divisible by $2$, a number is divisible by $3$ if the sum of its digits is divisible by $3$ (same rule for $9$ by the way). 
A number is divisible by $2^{n}$ if its last $n$ digits form a number divisible by $2^{n}$ (you can use this for 4 and 8) etc.
You have the same criteria as above for $5^{n}$ divisibility.
A number $ab...cd$ is divisible by $7$ if $ab...c-2d$ is divisible by $7$
Etc.
A: The number
$$N=a_0+10a_1+10^2a_2+10^3a_3+\cdots$$
is divisible by $7$ if and only if the number
$$a_0+3a_1+2a_2-a_3-3a_4-2a_5+a_6+3a_7+2a_8-a_9-3a_{10}-2a_{11}+a_{12}+\cdots$$
is divisible by $7$, and it's divisible by $13$ if and only if
$$a_0-3a_1-4a_2-a_3+3a_4+4a_5+a_6-3a_7-4a_8-a_9+3a_{10}+4a_{11}+a_{12}-\cdots$$
is divisible by $13$.
A: Another way of looking at the subtract twice the last removed digit technique others have described for 7 is that it's equivalent to dividing by 50, and seeing whether the remainder and quotient add up to a known multiple of 7.
Let's try this with 11172686, as you suggested.
Diving that by 50, you get 223453, remainder 36. Add the quotient and the remainder together, and you get 223489. Not sure whether that's a multiple of 7? Then we repeat the process with this new answer.
223489 divided by 50 equals 4469 remainder 39. The quotient and the remainder add up to 4508. Still not sure? Let's try it again.
4508 divided by 50 is 90 remainder 8. Quotient plus remainder gives us 98, and you should know that this is divisible by 7.
Why does this work? It's described in more detail here. That link also generalizes the technique for divisibility by any number ending in 9, or any prime number which can be scaled up to end in 9 (such as 7, which can be scaled up to 49).
A: $7 \cdot 11 \cdot 13 = 1001$
So, first its a good idea to check the remainder you get after dividing by $1001$. This will reduce your problem to examining a number with at most three digits.
11172686
From the right, break into groups of 3 --> 11, 172, 686
Subtract the sum of the 1st, 3rd, 5th, ... groups from the sum of the 2nd, 4th, 6th... groups
(11+686) - (172) = 697 - 172 = 127
None of 7, 11,13 is a divisor of 127.
