posted
Yeah, I'm having trouble getting to sleep. As a note, some of this is pretty geeky. It's mostly only of interest to people who are in to that sort of thing, if anyone's interested in it at all. If other people are interested, I can just post the rules of how to do this without explanation somewhere else.
Anyway, I think many people know the 9 times trick with your fingers.
Basically, to find out what 9 times some number (let's call it n) equals, all you need to do is hold up all your fingers and then count over n fingers and then hold that finger down. The tens digit will be the number of fingers you're holding up in the direction you started from and the ones will be the fingers in the other direction. So, for 9 times 4, you count over 4 fingers, which gives you 3 and 6, or 36.
I got to thinking, that works because of the properties of 10, so proper application of number theory should give a method to figure out all the numbers this way. So I did. It's actually pretty easy.
I'll try to explain. Each time you multiply by 9, you are basically adding a 10 with 1 taken away. This is captured by the finger trick because each step advances you one tens digit but removes one ones digit.
With the other numbers, it's a bit more complicated. Let's take 8. Now, you are adding 10 with 2 taken away. You need to account for the extra ones digit. So you need to add an extra step. You do the first step as with the 9s trick, but you also need to count one over from the other side. I'm going to assume that the first step happens from the right and the second from the left.
So 8 times 1, you drop one on the right and one on the left. Your hands look like _|||| ||||_. You've got 8 fingers up.
but now you run into a problem. For one thing, your dropping of fingers are about to criss cross. For another 8*6 = 48, which would take 12 fingers to represent. Also, our pattern from the first step expects that we'd have a 5 in the tens digit. But number theory comes to the rescue again.
When your finger droppings cross each other, that actually indicates that you're not going up a tens digit. The numbers you've taken away from each 10 you add (in the case of 8, 2) have exceeded 10. So at 8*6, you're taking away 12 from 60.
You can handle this with your fingers. But we need to add a third step (sort of). When your droppings cross, you just drop the finger to the right of the finger you dropped as part of the first step. There's actually a second part to this step, but let's deal with this first.
So, in the first step of 8*6 you count over 6 from the right and drop that finger: ||||_ |||||
The second step you drop 6 fingers from the left: _____ _||||
To get the 10s digit, we drop the finger to the right of the one from step 1. It's already down, but let's just pretend it wasn't for right now. This will give us 4 in the 10s digit. But how the heck do we get 8 in the ones digit?
Well, when we dropped the extra finger for the tens calculation, you could say we refreshed the ones digits. So, for the second part of the third step, put the fingers you put down in step 2, except for the one to the right of the one from step 1, back up.
Now your hands look like: ||||_ _||||
Haha. 44. Except wait, that's not right. That's because I've been hiding something about this till now. This is the last complicated bit though. When we were getting the ones digit, we weren't really just counting the fingers on the left side of the divide. It just looked that way because we hadn't crossed the divide. What we were actually doing was counting the number of fingers from where we ended up to the divide.
If you know that, you don't actually have to drop all the fingers for the second step. 8*2 could actually look like: |_||| |||_| ---6--- 1
Likewise, 8*3: ||_|| ||_|| --4-- 2
So, back to 8*6. We can either consider this the old way, with dropping all the fingers on the second step and then sproinging them back up as the second part of the third or use the new method. (I'm going to represent where you stop on the 2nd step as ! from now on - except in this one because we actually drop that finger as part of step 3.) Either way, our hands still look like: ||||_ _||||
The tens digit we've already gotten as 4. The ones digit is now all the fingers to the right of where we stopped to the divide or: ||||_ _|||| --4- and --4- because we wrap around.
This works for all the numbers down to 1 if you add one finger from the left for each number down (2 for 7, 3 for 6, etc.)
Posts: 10177 | Registered: Apr 2001
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posted
Incidentally, this is a somewhat modified version of the system I used to teach my 3 1/2 year old neice to multiply. She's scary smart, but she's also 3 1/2 and not a math prodigy.
So, given the interest in pedagogy lately, I guess I'm offering this up as an alternative to rote times table memorization. It's got the added bonus that, besides being sort of a game, I'm planning on using it to teach her about number theory.
Posts: 10177 | Registered: Apr 2001
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When I can't sleep, I count fields in increasing order. That is "There's a field of order 2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,36,37,41,43,47,49, etc)" Usually I fall asleep in the 60's ;-)
Posts: 168 | Registered: Jul 2006
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posted
I remember trying to figure out how to multiply on my fingers in different bases...don't remember if I succeeded, but I bet I frightened a lot of people in the cafeteria lunch line.
Then of course there's counting to 1,023 on your fingers when you're bored...and doubling when you're bored (this was before I'd read Ender's Game, by the way, and a way to pass the time in the bathtub)...and trying to come up with the general form for the "start with one coin on day one, and multiply it by x each day; what is the sum of all the coins on the nth day?" (You're a math geek, you figure it out on your own.)
I had an argument with my grandmother last week about doubling money. She was convinced that the only way to find out how much money you have on any given day (year, square, whatever) was to consult some kind of table. She simply refused to believe that I could possibly figure it out in my head based on an equation I'd learned.
I'm too young to have been taught new math, but I read a book called The Magic House of Numbers or something like that which had a lot of interesting math concepts, and figured stuff out for myself if it wasn't in any books I'd read.
Posts: 283 | Registered: Jul 2006
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posted
There is an easy way to multiply from 6*6 through 9*9 on your fingers...
Hold your palms facing you with fingers spread. Label little fingers 6, rings 7, middles 8, pointers 9. Thumbs don't get a label, but you count them for answers.
So, you put the two fingers representing the numbers that you want to multiply together tip to tip. The first part of the answer is the number of fingers including the touching ones and below multiplied by 10. The second part is found by multiplying the number of fingers higher than thje ones touching on each hand. The two are added together to find the total answer.
For example:
7 times 8 Put the tips of one ring finger (7) and one middle finger (8) together. There are 5 fingers touching or below, so 50. Above them are 3 fingers and two fingers, 3*2=6. So 7*8=56.
6 times 6 Put two little fingers together. 2 touching = 20. Four above on one hand times four above on the other hand is 16, so 36 total. 6*6=36
It may seem complicated, but with a little practice, it works.
Posts: 1379 | Registered: Feb 2002
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quote:Originally posted by Gwen: Then of course there's counting to 1,023 on your fingers when you're bored...and doubling when you're bored (this was before I'd read Ender's Game, by the way, and a way to pass the time in the bathtub)...and trying to come up with the general form for the "start with one coin on day one, and multiply it by x each day; what is the sum of all the coins on the nth day?" (You're a math geek, you figure it out on your own.)
Hey! I'm a math geek. Suppose I start with A coins (not 1 coin, neccesarily!), and every day I multiply it by x. How much do I have on the nth day?
That's easy! A*(1-X^(n+1))/(1-X)
And, if X is between -1 and 1, then I can actually let the process continue infinitely (assuming the dollar infinitely divisible...which it's not), and the total amount of money I'll have is A/(1-X)!
posted
Taking square roots by hand (using 178929 = 423^2 as an example:
1: Write out the number in two digit increments, starting with the ones (17 82 29). Put a radical over it, sort of like you were doing long division.
2: For the leftmost pair, find the highest digit with a square less than the pair (4^2 = 16, here). Write the digit above the square root line, and its square below the pair, subtracting and leaving a remainder as if you were doing long division (17-16 = 1).
3: Drop the next pair (189)
4: Take what you have, double it, and put a _ next to it under the number (8_, here). Look for the largest digit that will fit in the blank so that when the full number is multiplied by the digit, it's less than the remainder from step 3. Here, that's 2: 82 * 2 = 164 < 189, 83 * 3 = 249 > 189.
5: Write the digit above the radical, subtract the multiplied number from the remainder in step 3 to get a new remainder. (Write 2 above the radical next to the 4, 164 under the 189, subtract to get 25).
Repeat steps 3-5 until you pull out as many decimal places as you want. We'll draw out another step:
Drop down the next pair to get 2529 42*2 = 84 Look for the digit to fill in 84_. 843*3 = 2529. Write 3 above the radical, and subtract to get a remainder of 0, showing that we started with a perfect square.
Hopefully that makes sense?
Posts: 170 | Registered: Mar 2001
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I've got a different trick, works kinda backwards.
178929 is close to 160000 = 400^2, leaving 18929.
Divide that by 400 = 47 with a remainder of 129.
Divide 47 in half. It gives you 23 with a remainder of 1.
Add 129 to the remainder (1) * 400. 529. Square 23 = 529.
So, your answer is exactly 423.
This works because this difference between the square of n and that of n + x = n(x) + (n + x)(x) = (2n + x)(x) so 423^2 = 160000 + 400 * 23 + 423 * 23 = 160000 + 2 * (400 * 23) + 23 * 23
posted
ps - I'm good at everyday math... but you guys are waaaaaaay better at MATH (and it's cool tricks) than me... Nice Work!
Posts: 1355 | Registered: Jul 2006
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