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It has just occurred to me that there is a very simple test to check if an integer is divisible by 4: take twice its tens place and add it to its ones place. If that number is divisible by 4, so is the original number.

This result seems like something that anybody with an elementary knowledge of modular arithmetic could realize, but I have noticed that it is conspicuously missing on many lists of divisibility tests (for example, see here, here, here, or here). Is this divisibility test well-known?

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  • $\begingroup$ I would say that it's reasonably well-known; it's also equivalent (of course) to 'if the tens place is odd, then the number is divisible if the last digit is even and not divisible by $4$ (i.e., $2$ or $6$); if the tens place is odd, then the number is divisible if the ones place is (i.e., $0$, $4$ or $8$)', which is the version that I seem to use mentally (since it just involves odd-or-evenness and a single digit test, which I can do even more quickly than a double-and-add). $\endgroup$ – Steven Stadnicki Oct 17 '13 at 16:05
  • $\begingroup$ Well, it is there at en.wikipedia.org/wiki/Divisibility_rule Probably not so well used compared to other ways. $\endgroup$ – Macavity Oct 17 '13 at 18:10
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Yes, it is well known; do you know modular arithmetic? Assuming you do, we have a number $abc=a\cdot 10^2+b\cdot 10^1+c\cdot 10^0$. Now $$a\cdot 10^2+b\cdot 10^1+c\cdot 10^0\equiv 2\cdot b+c\pmod{4}.$$ Many people know the multiples of $4$ for numbers less than $100$, so it is commonly just said if the last two digits (as a number) is divisible by $4$, then the number is divisible by $4$.

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  • $\begingroup$ Yes, that's how I came to realize it. It seems like such a simple result, but I noticed that it is conspicuously absent in many places. $\endgroup$ – David Zhang Oct 17 '13 at 16:07

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