The Universe of Discourse

Thu, 20 Jan 2022

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.

But one reader was surprised that I called it “useless”, saying:

I only know of one test for divisibility by 8: if the last three digits of a number are divisible by 8, so is the original number. Fine … until the last three digits are something like 696.

Most of these divisibility tricks are of limited usefulness, because they are not less effort than short division, which takes care of the general problem. I discussed short division in the first article in this series with this example:

Suppose you want to see if 1234 is divisible by 7. It's 1200-something, so take away 700, which leaves 500-something. 500-what? 530-something. So take away 490, leaving 40-something. 40-what? 44. Now take away 42, leaving 2. That's not 0, so 1234 is not divisible by 7.

For a number like 696, take away 640, leaving 56. 56 is divisible by 8, so 696 is also. Suppose we were doing 996 instead? From 996 take away 800 leaving 196, and then take away 160 leaving 36, which is not divisible by 8. For divisibility by 8 you can ignore all but the last 3 digits but it works quite well for other small divisors, even when the dividend is large.

This not not what I usually do myself, though. My own method is a bit hard to describe but I will try. The number has the form !!ABB!! where !!BB!! is a multiple of 4, or else we would not be checking it in the first place. The !!BB!! part has a ⸢parity⸣, it is either an even multiple of 4 (that is, a multiple of 8) or an odd multiple of 4 (otherwise). This ⸢parity⸣ must match the (ordinary) parity of !!A!!. !!ABB!! is divisible by 8 if and only if the parities match. For example, 104 is divisible by 8 because both parts are ⸢odd⸣. Similarly 696 where both parts are ⸢even⸣. But 852 is not divisible by 8, because the 8 is even but the 52 is ⸢odd⸣.

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