Kevin Fagin’s Drabble from Sunday poses a nice bit of recreational mathematics, the sort of thing one might do for amusement: Ralph Drabble tries to figure how long he’s spent waiting at one traffic light. I want to talk about some of the mental arithmetic tricks I’d use to get through the puzzle without missing the light’s change. In the spirit of the thing I’m doing the calculations for this only in my head, though I admit checking with a calculator afterward to see if I got close.

Mental arithmetic is, with the apparent exception of a few shockingly talented people (whom I’ve read were in high demand among astronomers and physicists in the days before mechanical calculators), a matter of approximation and simplification: how can you get something tolerably close to the answer you want while doing problems that are easy enough to do in your head?

Ralph Drabble starts off well here by estimating the traffic signal to last about 90 seconds. Obviously it’s not 90 seconds every time, but it’s reasonable to suppose that the occasions he has very short waits are balanced by times he has longer waits than usual. In any case replacing the whole spectrum of possible waits with one time is a good move, particularly since he chose an easy number. 90 isn’t very easy in general, but it *is* one and a half minutes, and since he figures he passes the light twice a day, that’s 3 minutes total per day at the light. 3 is a very easy number to work with, or at least much easier than 90.

He also supposes he passes the light twice every single day, 365 days per year. This is another good simplification. Surely there are days he doesn’t leave the house, or is on vacation, or so; but there are surely days he has to make several trips away from home. It would be a bit surprising if they balanced to exactly one pair of visits per day for an entire year, but it’s much easier to work with 1 pair of crossings than, say, a more precise 1.224 or whatever the actual average might be. Also his approximating a year as 365 rather than 365-and-a-bit-under-a-quarter days is wise. Given that we’ve already approximated time spent at the traffic light per day at 3 minutes, rather than, say, 2.85 minutes, it’s pointless to estimate the year more exactly than that.

He goes wrong in the next step, in figuring that as it’s 3 minutes a day, times 365 days per year, times 20 years, that he then needs to multiply 3 by 365 by 20. That’s not truly wrong, no, since that is the calculation to do. But it’s a bad approach to use for mental arithmetic since 365 is an *awful* number to multiply by. I’d refuse to multiply 365 by 3 in my head. If I *had* to, I would instead note that 365 is 400 minus 35. Then, by the distributive property, 365 times 3 must be 400 times 3 minus 35 times 3, or 1200 minus 105, or 1095. I hope. Distribution is one of mental arithmetic’s greatest powers. It effectively lets you turn multiplication by a multiple-digit number into a set of multiplications of single-digit numbers, at least if you’re willing to believe that 400 times 3 is no harder than 4 times 3, with some addition or subtraction on the side. It’s also good for noticing that some number is close to, say, one-third of a hundred, or two-fifth of ten thousand, or so.

(The other way I might multiply 365 by something, if I had to: it strikes me that 365 is pretty near to 333 plus 33. That looks worse, except, 333 times 3 is just about 1000, and 33 times ~~33~~ 3 is just about 100. If I know I’ll be multiplying 365 by some multiple of three, I can leap directly to 1100 and not be too far off. I suppose this spoils my claim that he should have multiplied 3 by 20 first, but it’s pretty clear he didn’t see what 3 times 365 might be.)

But here’s the right way to figure it: remember that multiplication commutes. We can change the order of the things we’re multiplying in any way we like. So 3 minutes times 365 days-per-year times 20 years is the same thing as 3 times 20 times 365. And, ah, 3 times 20 is 60. 60 is nice to have here because 60 minutes is one hour, so we can change that 60 to 1, and get a result not in total minutes spent at the light but instead total hours spent at the light. And 1 times 365 is easy enough you don’t even need to do it in your head. He’s spent something like 365 hours at the light. Commutation is another great tool for mental arithmetic, and it works in addition as well as multiplication, so if you do much of it you’ll pause before calculating to figure whether swapping some multiples around might give you an easier problem. Often it does.

Granted, nobody has much idea how long 365 hours is. But there are 24 hours in a day, and 24 goes into 365 … well, that’s a pain. But 365 is pretty near 360 and everything divides into 360. I exaggerate but not by much; it’s likely that the great number of things that divide evenly into 360 is part of why we have 360 degrees in a circle, and so many early calendars set the year at 360 days plus a handful of intercalary days.

24 going into 360 is a nice, easy proposition: that’s 15. So there were 15 days spent at the traffic light, which sounds like a lot until you remember that’s spread out over 20 years, which over 365 days per year over 20 years comes out to about three minutes a day, not very long at all.

PS: I got the calculations right. Of course, I would say that. Or I used the calculator wrong.

I think you wrote “33 times 33” instead of “33 times 3”. Nice article nevertheless :-).

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Oh, blast, you’re right. That ought to have been 33 times 3.

Thank you.

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Mathieu says:

Thursday, 22 March, 2012 at 9:46 am

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I think you wrote “33 times 33” instead of “33 times 3”. Nice article nevertheless :-).

Sally Brown was correct, you never can trust 3s

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Argh and yes, there’s either one or ten times too many 3’s in that sentence. Thanks for catching it.

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