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Thu, 06 Jan 2022

Another test for divisibility by 7

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Recently I thought of another way to check for divisibility by !!7!!. Let's consider !!\color{darkblue}{3269}!!. The rule is: take the current total (initially 0), triple it, and add the next digit to the right. So here we do:

$$ \begin{align} \color{darkblue}{3}·3 & + \color{darkblue}{2} & = && \color{darkred}{11} \\ \color{darkred}{11}·3 & + \color{darkblue}{6} & = && \color{darkred}{39} \\ \color{darkred}{39}·3 & + \color{darkblue}{9} & = && \color{darkred}{126} \\ \end{align} $$

and the final number, !!\color{darkred}{126} !!, is a multiple of !!7!! if and only if the initial one was. If you're not sure about !!126!! you can check it the same way:

$$ \begin{align} \color{darkblue}{1} ·3 & + \color{darkblue}{2} & = && \color{darkred}{5} \\ \color{darkred}{5} ·3 & + \color{darkblue}{6} & = && \color{darkred}{21} \\ \end{align} $$

If you're not sure about !!\color{darkred}{21} !!, you calculate !!2·3+1=7!! and if you're not sure about !!7!!, I can't help.

You can simplify the arithmetic by reducing everything mod !!7!! whenever it gets inconvenient, so checking !!3269!! really looks like this:

$$ \begin{align} \color{darkblue}{3} ·3 & + \color{darkblue}{2} & = && 11 = \color{darkred}{4} \\ \color{darkred}{4} ·3 & + \color{darkblue}{6} & = && 18 = \color{darkred}{4} \\ \color{darkred}{4} ·3 & + \color{darkblue}{9} & = && 21 = \color{darkred}{0} \\ \end{align} $$

This is actually practical.

I'm so confident this is already in the Wikipedia article about divisibility testing that I didn't bother to actually check. But I did check the email that Eric Roode sent me in 2018 about divisibility testing, and confirmed that it was in there.

Instead of multiplying the total by 3 at each step, you can multiply it by 2, which gives you a (correct but useless) test for divisibility by 8. Or you can multiply it by 1, which gives you the usual (correct and useful) test for divisibility by 9. Or you can multiply it by 0, which gives you a slightly silly (but correct) version of the usual test for divisibility by 10. Or you can multiply it by -1, which which gives you exactly the usual test for divisibility by 11.

You can of course push it farther in either direction, but none of the results seems particularly interesting as a practical divisibility test.

I wish I had known about this as a kid, though, because I would probably have been interested to discover that the pattern continues to work: if at each step you multiply by !!k!!, you get a test for divisibility by !!10-k!!. Sure, you can take !!k=9!! or !!k=10!! if you like, go right ahead, it still works.

And if you do it for base-!!r!! numerals, you get a test for divisibility by !!r-k!!, so this is a sort of universal key to divisibility tests. In base 16, the triple-and-add method tests not for divisibility by 7 but for divisibilty by 13. If you want to test for divisibility by !!7!! you can use double-and-add instead, which is a nice wrinkle.

The tests you get aren't in general any easier than just doing short division, of course. At least they are easy to remember!

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