Math Geekery

I’m pretty good with my divisible-by tests. Let’s see, off the top of my head…

If all the digits added together are divisible by 3, then the number is divisible by three. So, 1,493,221,848 is divisible by 3 because 1+4+9+3+2+2+1+8+4+8 = 42, and 4+2 = 6, which is divisible by three.

Likewise it’s divisible by 6, because it’s divisible by 3 and it’s an even number.

And so on. There are easy tests for numbers up to 11, except for 7. Here’s a not-easy test:

Steve Landry Facebooked this (is that a word?), pointing to Tanya Khovanova’s Math Blog. Here’s how it works, per Tanya:

Write down a number n. Start at the small white node at the bottom of the graph. For each digit d in n, follow d black arrows in a succession, and as you move from one digit to the next, follow 1 white arrow.

For example, if n = 325, follow 3 black arrows, then 1 white arrow, then 2 black arrows, then 1 white arrow, and finally 5 black arrows.

If you end up back at the white node, n is divisible by 7.

Cool, eh? But unlike the rules I’ve memorized, I don’t think this one’s going to stick with me.

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