# What is the point of the common divisibility trick for $7$?

The "divisibility rule" to test whether a given integer is divisible by $7$ (or, more generally, to find the remainder when an integer is divided by $7$) is in my opinion, ridiculous. The method is so convoluted that I can't even remember it off the top of my head, and it is almost certainly slower than simply subtracting easy multiples of seven. For example:

\begin{align} 1234&\,\\ \color{red}{-700}&\,\\ 534&\,\\ \color{red}{-490}&\,\\ 44&\,\\ \color{red}{-42}&\,\\ 2&\,\\ \end{align}

Another one:

\begin{align} 314159&\,\\ \color{red}{-280000}&\,\\ 34159&\,\\ \color{red}{-28000}&\,\\ 6159&\,\\ \color{red}{-5600}&\,\\ 559&\,\\ \color{red}{-560}&\,\\ -1&\, (6) \end{align}

This method seems so much better, but is there some merit to the divisibility "rule"? The specific one I'm thinking of is the first one mentioned here (I will admit that the Pohlman-Mass method seems interesting).

• I agree; my preferred method is also to subtract multiples, at least for mental arithmetic. – Qiaochu Yuan Mar 12 '15 at 6:28
• The Wikipedia page seems to be quite a grab-bag of divisibility methods. Some of them are mentioned once and then mentioned much later on the page without referring back to the original description. The quality of explanation of methods is also somewhat variable. I think what you are experiencing is a lack of appropriate editing of that page (possibly including a poor choice of which method to list first). – David K Mar 12 '15 at 7:06

I ran through your two examples using the method you linked (assuming you're talking about the $10x+y$, $x-2y$ method).

1234
123-2*4=115
11-2*5=1

314159
31415-2*9=31397
3139-2*7=3125
312-2*5=302
30-2*2=26

So there are 2 steps in the first instance and 4 steps in the second. This is equal to or less than the number your method took (assuming you can recognise easily that 26 is not a multiple of 7). So it's possible that the number of steps is comparable. It's certainly bounded by the number of digits in the number since each step effectively divides by 10.

The second thing I would note is that your method requires knowledge of mulitiples of 7 where as the linked method only requires knowledge of subtraction and how to multiple a single digit number by two.

In fact, in the linked method, all the information you need to know to solve the problem is contained in the number and is readily accessible. I'd argue that your method requires information that is (slightly) hidden. And the subtractions you make are by no means unique which makes it somewhat less formulaic.

Finally (and slightly tangentially) I would add that the linked method certainly has a curiosity value, but may in fact be useful for proving a theorem in numer theory for example - not that I know of any.

I probably would use your method if I had to find determine if a number was a multiple of 7, but the linked method does have some value and some people may actually prefer it. Really, it comes down to different horses for different courses.

• I think this is a better explanation than Wikipedia's for this particular technique. – David K Mar 12 '15 at 7:09

A number of the form $$10a+b$$ is divisible by $$7$$ if and only if $$a-2b$$ is divisible by $$7.$$

I do not have any difficult memorize this rule because of this single line proof of the test: $$2(10a+b)+(a-2b)=21a.$$

• Are $a,b$ assumed to be digits, or any numbers? – Allawonder Oct 25 '19 at 4:25
• @Allawonder: Let me give you an example. For $1234,$ here $a=123$ and $b=4.$ – Bumblebee Oct 25 '19 at 4:36
• Alright, thanks. My question's answered. – Allawonder Oct 26 '19 at 16:07