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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.

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  • $\begingroup$ use that $7\cdot 11\cdot 13=1001$ $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '14 at 9:21
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    $\begingroup$ Use a rule for the digit sum. $\endgroup$ – Dietrich Burde Oct 23 '14 at 9:21
  • $\begingroup$ @DietrichBurde Thanks for the Wiki link but it seems too complicated for me $\endgroup$ – L Lawliet Oct 23 '14 at 9:35
  • $\begingroup$ I have asked a very similar question a while ago (math.stackexchange.com/q/918317/131263), and got a pretty good answer for it. $\endgroup$ – barak manos Oct 23 '14 at 9:47
  • $\begingroup$ @barakmanos Thank you for the link $\endgroup$ – L Lawliet Oct 23 '14 at 9:55
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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 $13$

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  • $\begingroup$ Can you elaborate your method more detail? It seems work. Thanks $\endgroup$ – L Lawliet Oct 23 '14 at 9:26
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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.

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Specifically for divisions with $7$ and $13$ (and $11$) there are very simple tests available online.

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!

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!

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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 $13$.

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  • $\begingroup$ Thanks for your answer, +1. Do you know what are the rules for the other numbers? $\endgroup$ – L Lawliet Oct 23 '14 at 10:17
  • $\begingroup$ The rules are all derived in the same way. Are you familiar with congruence notation (modular arithmetic), i.e., do you know what $100\equiv2\pmod7$ means? $\endgroup$ – bof Oct 23 '14 at 10:29
  • $\begingroup$ No. What does that mean? $\endgroup$ – L Lawliet Oct 23 '14 at 10:32
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    $\begingroup$ $a\equiv b\pmod{m}$ means that the difference $a-b$ is divisible by $m$; in other words, dividing $a$ or $b$ by $m$ leaves the same remainder. So $10\equiv3\pmod7$, $100\equiv2\pmod7$, $1000\equiv-1\pmod7$, etc. This is why, in the number $a+10b+100c+1000d$ we can replace $10$ with $3$, $100$ with $2$, etc., without affecting divisibility by $7$. Divisibility by $11$ is an easy case because $10\equiv-1\pmod{11}$, so the coefficients are alternately $+1$ and $-1$. Divisibility by $9$ is even easier, since $10\equiv1\pmod9$. $\endgroup$ – bof Oct 23 '14 at 10:48
  • $\begingroup$ Thanks for the long explanation. I'll try to learn this topic and add it to my math toolbox $\endgroup$ – L Lawliet Oct 24 '14 at 12:47
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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 $13$

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  • $\begingroup$ it is the same rule since $7\cdot 11\cdot 13=1001$ $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '14 at 9:50
  • $\begingroup$ your number is also not divisible by $11$ $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '14 at 9:51
  • $\begingroup$ you have to calculate $313-1=312=24\cdot 13$ $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '14 at 10:30
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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).

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