June 2023: Recovering My Mental Mathematics

Over the last few years I've somewhat given up on practicing solving equations on the fly. It's not that I'm particularly bad at it, but rather that I just never needed it much, at least not beyond simple addition/multiplication. But since it looks like I'll need it more going forward it might be worth thinking about it more now. However, since I've never been great at it, and there's not a good way to shoot beyond that ceiling, I decided to pick a fingercounting system to learn that will enable if not faster computation, more flexible ones.

In Europe, we mainly learn to count with the ten fingers we have, which can get us to 100 comfortably, if we use ASL counting, but isn't great for actual calculation. Turns out, those stories are really only the start of the story. There isn't really one system of mental and finger mathematics. My problem with most mental arithmetic is that I am bad at remembering things. I like to joke that I have the short-term memory of a goldfish. However, I think that might not a problem unique to me, at least not when it comes to mental arithmetic. So let's go by the order of easiest, and start with addition (and subtraction, I'm still an abstract algebra fan by heart).

Chisanbop

There is a Korean method that outsources addition and subtraction to bodyparts that aren't your head. The one resource I see popping up everywhere is The Complete Book of Chisanbop, which isn't too easy to get your hands on where I live, unless you want to lose an inordinate amount of money in exchange for being able to add numbers together with your fingers

So I did the sensible, unmentionable thing, people do on the internet. The concept is actually quite easy. It divides the digits of a number onto the hands, where the ones go on the fingers and the fives go on the thumbs. Starting out, the right hand tracks the number 1 - 9 left tracks 10 - 90 in steps of ten. Then, addition or subtraction can be done digit-wise, without actively needing to do calculations. You can basically read the result off the hands when you're done. Subtraction works basically the same. However, I'm greedy and I want to have more than 99 numbers. Luckily there's a way to extend the system to 9999 by half-extending fingers on both hands, which effectively doubles the "number of fingers" you have to work with, and honestly, I don't think I'll have to do more than that. Practicing this is going to be really boring, because it's just associating the numbers with finger-positions, and doing actual calculations feels like abstract piano playing, but it enables the mental calculations of really long sums.

Multiplication

This is the part where it gets fun, if slightly disappointing. From what I could find, there isn't really any way to do this on a pair of hands that doesn't get tedious once we get to double digit multipliers. It's also not one system for multiplication either, but rather an ensemble of tricks that exploits some oddities in certain numbers. There's some methods to do single-digit calculations on the hands similar to how Chisanbop does additions, but that's not anything I'm too interested in learning, since those particular problems have been successfully drilled into my head.

At the time of writing it's the 30th of June in the evening and I'm just trying to recall the most memorable ones. One of the easiest is probably any number n multiplied by eleven, where we take the first and last digits of n as the first and last digits of the result, then add each neighboring digits of n to get the digits in between.

Multiplying any two-digit numbers whose first digits are the same and whose last digits add up to ten is done by multiplying the first digit with (itself + 1) to get the first two digits of the result, and the multiply the last digits to get the last two digits of the result.

For any other two digit numbers that have the same first digit, add the unit digit of one to the other number, and multiply this by the shared first digit. Multiply the whole thing by 10 to get the correct number of digits and add the product of both unit digits.

Division

Unfortunately inverting multiplication isn't quite as easy as inverting sum. Essentially, the decimal representation of each 1/n fraction can be memorized (there's only 9 of them), since they are either nice fractions, or repeating fractions. The rest is multiplication by this representation. This process is highly iterative, but it's much better than not having anything at all and having to bust out the pen and paper. I'm pretty sure this one will require the most practice, as far as I'm concerned, but that's fine.

Square Roots

You need to know a few things to solve square roots in your head. It doesn't require you to do stuff with fingers, but on the bright side, you only need to know the multiplication table up to 100, which isn't a lot. It comes from the perfect squares, specifically the last digit of the perfect squares form 1 to 9.

Any 4-digit perfect square can be split into two 2-digit numbers. Because of how multiplication and digits work, we know that a 4-digit (and also every 3-digit, I guess) perfect square has exactly 2 digits. We need to check what the first digit can be, so we take the next-smallest one of the left 2-digit number. That's just knowing your tables, so that should be trivial. Then we multiply the number n we settled on and calculate n(n+1). Let's call it m. Now we check the last digit of the right 2-digit number, which gives us 2 possible options. If the right 2-digit number is larger or equal than m, then we take the larger of the two, otherwise we take the lower.

Any further and you'll just kinda need to know your squares beyond 100, which I never bothered to memorize. Goldfish-brain, remember? I might do this eventually though, since every digit of squares you can add, will net you 2 more digits with this method.

If you're so inclined, you could try and guess the next perfect square for 4-digit numbers, which can be verified by rounding the number to the next 10 (call this n). Apply the inverse operation to that nearest 10 twice (call this m). Note the squared difference between the number to square and the nearest 10 as s. The result will be n*m + s.

For non-perfect squares, one does actually require the perfect square below it, for which the above is prerequisite. If we know that number, it's really just a number of getting the decimal points, and the best I could find there was a decent approximation: take the difference between the perfect square and the number in question, then divide it by double the root. This will give a fraction, which we should be able to add to the result.

There's books upon books on this topic. One I've found recommended a couple times was Secrets of Mental Math. I haven't looked into this yet, so I can't say much on its contents. For now I'm good and satiated with these tricks, especially the multiplication ones. The good thing is, that I run into these trivial math problems every now and then, and these are things you can practice on the spot, provided you can remember the tricks. I, for one, will try and keep those in mind, and maybe I'll take a look at a book or two once these methods have become routine.

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August 2023: Forcing Myself to Try Digital Art