Sat, 21 Nov 2020
[ Previously, Testing for divisibility by 7. ]
A couple of nights ago I was keeping Katara company while she revised an essay on The Scarlet Letter (ugh) and to pass the time one of the things I did was tinker with the tests for divisibility rules by 9 and 11. In the course of this I discovered the following method for divisibility by 19:
The result will be a smaller number which is a multiple of 19 if and only if the original number was.
For example, let's consider, oh, I don't know, 2337. We calculate:
76 is a multiple of 19, so 2337 was also. But if you're not sure about 76 you can compute 2·6+7 = 19 and if you're not sure about that you need more help than I can provide.
I don't claim this is especially practical, but it is fun, not completely unworkable, and I hadn't seen anything like it before. You can save a lot of trouble by reducing the intermediate values mod 19 when needed. In the example above above, after the first step you get to 17, which you can reduce mod 19 to -2, and then the next step is -2·2+3 = -1, and the final step is -1·2+2 = 0.
Last time I wrote about this Eric Roode sent me a whole compendium of divisibility tests, including one for divisibility by 19. It's a little like mine, but in reverse: group the digits in pairs, left to right; multiply each pair by 5 and then add the next pair. Here's 2337 again:
Again you can save a lot trouble by reducing mod 19 before the multiplication. So instead of the first step being 23·5 + 37 you can reduce the 23·5 to 4·5 = 20 and then add the 37 to get 57 right away.
[ Addendum: Of course this was discovered long ago, and in fact Wikipedia mentions it. ]
[ Addendum 20201123: An earlier version of this article claimed that the double-and-add step I described preserves the mod-19 residue. It does not, of course; the doubling step doubles it. It is, however, true that it is zero afterward if and only if it was zero before. ]