EPT-2/1 added part 3
-2/2 added part 4
-2/2 added Addendum
I posted this as a note on Facebook, but I found an old thread that didn't have much so I figured I'd share it here too.
Ever since I was a child, I've been obsessed with doing arithmetic in my head. When I learned about the abacus, I begged my mom to sign me up instantly. The abacus arithmetic uses very similar princples as the ones I'm about to show you, except the difference is it is done in such a way that it makes mental calculations incredibly efficient, and memorable. So I took lessons, then moved to the United States of America, where I didn't have an abacus tutor anymore, and I learned conventional arithmetic, which I had always thought was boring, inefficient, and that it could not be the way that the math whizzes did large multiplications in their head. Most people think that they are unable to do mental arithmetic because of the brain capacity, but it's not; it is the system that has reached its limits, not the brain. The brain is capable of handling a lot. With a bit of practice and a good system, anything is possible.
Recently, mainly due to boredom in my Poli Sci/Number Theory classes, I've worked out a series of things from numerous sources and my own experimentation that has led to some insane mental calculation skills [which I am by no means a master, but I'm working on it]. And now I shall reveal them to you.
Prologue: Why do mental arithmetic?
It's so easy to whip out a calculator and just have that do your calculations. And nowadays, phones have calculators too. But there are some situations where you cannot use calculators; for example, playing poker [if you haven't memorized your card odds], you don't have your phone on you, or you simply want to know what 3466^2 is [which is 12013156] for whatever reason.
But there are much better reasons. Mainly because it trains your brain, and enables you to keep track of more at the same time, and is just a good brain workout in general. I'm sure it'll do things like prevent Alzheimer's.
Prerequisites:
-The ability to perform any multiplication with the digits from 1-9 almost instantaneously
-"Carry spotting," the same as the previous except you only tell me the first digit. Ex: 6 x 4 -> 2, 9 x 9 -> 8, etc
-The ability to perform any addition with the digits from 1-9 almost instantaneously (we will extend this...by a lot)
-The ability to subtract any digits from 1-9 instantly.
-Memorization of the squares from 1-30 (not necessarily THAT required, but helpful)
And that's really all you need. Everything else is developed.
Notation:
/ represents skewed addition, or adding the numbers on either side of the slash. For example, 4/7 = 47, 654/35 = 6575.
// represents double digit skewed addition, so adding the two digits on either side of the double slash. So 463//225 = 4855.
And so on...
1. Two Digit Addition
We're going to start really simple and then work our way up to the hard stuff. Two digit addition is really simple, if you recognize that you have to work from left to right, as opposed to from right to left. Left to right is much more efficient, and I have no idea why they don't teach it.
For example, 84 + 17. I instantly call out 101 because I've done so many of these, but what really happened was, after I looked at the number, I realized the first digits were going to be 10, and that the last digit had to be 1.
So add 8 + 1 first to get 9. But this is in the tens digit, so what you really have is 90. Then add 4 and 7 to get 11. Add the two together to get 101. I know it sounds not much different than working from left to right, but it is more effective to work this way so you do not have to reverse your answer in your head.
This is relatively straightforward, but it allows me to introduce a concept I call "addition complement pairs," for lack of a better name, and this is something you should memorize too. The borderline additions get really awkward, so knowing them instantaneously (if you already know them, that's fine) is really helpful. I'll list the first few and let you complete the list.
5 , 5 -> 10
5, 6 -> 11
...
5, 9 -> 14
6, 5 -> 11 (identical to 5,6)
...
9,9 -> 18
You need to know these like the back of your hand, they show up throughout everything we do.
2. Two Digit Addition of Many Numbers
a. Able to see them
It makes a surprisingly huge difference between being able to see the numbers and not being able to see the numbers. I've gotten so good at it I just add the numbers as they are being given to me, but if I wasn't allowed to do that, that would be a huge hassle. But let's say I just had them out in front of me. That's really great.
So let's do 15 + 63 + 35 + 39 + 49 + 72. I just hit random numbers on my keypad and hoped for the best. So the fundamental technique of this is to look for groups that add up to 10 - and then tally it up and throw the rest away. You can work left/right or right/left to this, the result is the same; I prefer left->right for coherence purposes. So I notice 1, 6, 3 form a pair of 10, so I save 10 in my head, and I notice 3 and 7 make another, leaving 4. So we're at 240 so far. Looking at the ones digits, I notice 5 and 5 make a 10, 9 and 9 and 2 make a 20, and that leaves us with a 3, so we're at 33. Hence the answer is 273. If you have trouble remembering the numbers, I'll cover that in a later chapter when we deal with insane multiplications.
Do you see where pairing comes in? I had to recognize immediately that 9,9 make 18, and that 2 completes it. If I did not, this would have been much farther.
b. Unable to see them
This is much harder; you do not have the benefit of visualizing the numbers; I would recommend you develop the skill to add to a running total as you are fed this information. When I get to mnemonics (and consequently, mathematics with mnemonics), this will become significantly easier. So I can't really demonstrate this right now.
3. Concept: Trick Extension
Any trick that I discuss here can be extended infinitely, once you spot the pattern. For example, the same pattern I use to do two digit squares that begin with 5 work with 5 digit squares that begin with 5.
I can extend the two digit addition trick to 4 digits. So let's say I'm given a series of four digit numbers
2046 + 6542 + 7682 + 1875 + 1237 + 4556
So looking at the first digits; 2-7-1 make a pair, 6-4 another, and 1 left over, so that's 21, which is nut. So remember nut and keep look at the next digits. 0, yes so nice; 5-5 make a pair, 8-2 make a pair, and 6 remains, so 26. So we're now at 236 because 21(0) + 26 = 236 (all I did was move it over one digit and add), which, if you're doing the mnemonic technique, is "No Amish." So I picture a sign saying "no Amish" and I leave it at that. Next digits; 7-3 make a pair, how lovely, then we have that awkward 5, and 4-4-8, which together is 21, so we have 31. So 236(0) + 31 = 2391. Which becomes "gnome bite." Finally, the last digits: 6-2-2 make a pair, and the remaining digits make 18, so we have 28, and performing the skewed addition yields 2391(0) + 28 = 23938. (Gnome be moving) I will cover the mnemonics later and how to maths with them.
This might seem slower than just doing it by hand, but after a while, this stuff becomes natural, and your mind just speeds through these calculations. I'm not quite there with addition because I'm lazy and I don't practice it, but it's very possible.
4. Two Digit by One Digit Multiplication
I credit this trick from Arthur Benjamin's book, "Secrets of Mental Math" (which takes a slightly different approach than the ones I use here, though some of the tricks I use and there are some cool underlying concepts in there, I definitely recommend it), and we will later see how this relates to methods from Trachtenberg.
So most people, when they try to multiply things like 13 x 8, think that they have to do 3 x 8, which is 24, and then add 2 to 1x 8, which is 10, to yield 104. It may seem elementary for such small numbers, but what happens when you get 78 x 9? Then things get a little bit hairier. They'd think, 8 x 9 = 72, so carry the 7 and remember 2 (and sometimes they don't remember 2; I had this problem when I was younger), and then 7 x 9 = 63 + 7 is 70, so 702 is my answer.
well, we want to work left to right so we don't have to reverse our answer in order to spit it out. So we'll use "divide and conquer." This is why knowing your times table is essential.
78 x 9 = (70 + 8) x 9 = 70 x 9 + 8 x 9 = 630 + 72 = 702. This is much faster, especially if you think of it like this:
83 x 6. -> 48/18 -> 498. (combine the middle digits)
39 x 8 -> 24/72 -> 312. (why pairing is important)
Note: When you add two single digit numbers, the carry will NEVER be more than 1. NEVER. So if there's only 2 and you see you'll carry, just add 1 automatically to the digit before and choose the proper pairing.
5. Two Digit by Two Digit Multiplication; Two Digit Squares and Numbers close to 100
This is the fruit of this note by the way, since I'm getting tired and I want to do something else before I sleep, so I'll show you all the other tricks some other time. But this is good enough to get you started.
So the previous trick relied on breaking them down, and while you can do that for 2 digit x 2 digit, it's kind of annoying. Case in point:
74 x 69. So we'll do 74 x 9 first which is 666. (Verify this.) Then we'll do 74 x 6 which is 444. (Verify this as well.) Then we'll do a skewed addition. Which leaves us with 5106. Which wasn't that bad, but there are better ways.
First of all, notice that 69 is really close to 70. So that's 74 x 7? It's 518. So 74 x 70 = 5180, and if I take 74 away from that, I get 5106. (Because 74 x 69 = 74 x (70-1) ). So you can use this trick for numbers really close to an even number.
So another example. 64 x 99. Well, 99 is really close to 100, so I'll do 64 x 100 - 64 = 6400 - 64 = 6336. If you want to do things like "100 - x," just remember that unless the last two digits are zero, the first digits will always sum to 9 and the last digits to 10. These are the "complements to 100," (by Arthur Benjamin, though I discovered this concept independently in like 4th grade but he deserves giving it a name). So like 34 has the complement 66, since 3 + 6 = 9 and 4 + 6 = 10. I think you get the point. While I'm on this tangent; let's say you want to do something liek 143 - 84. Something no one likes to do. Well, what you can actually do is instead of subtracting by 84, you can subtract 100 and add 16 (since 16 is the complement of 84, as 100 - 16 = 84, so 143 - 84 = 143 - 100 + 16 = 59.)
These are great, but what about things not that close to anything? Like 63 x 96? (Ew, that's ugly). Well, introducing cross-multiplication! (This can be extended, and this is how I do mass multiplication, as I haven't discovered a better method for this yet)
So I'll do it first then explain. Now 96 is really close to 100, but 63 x 4 is annoying and I really don't want to do that, much less subtract that from 6300. Though by the time I finished typing this sentence I realized that it was 6048, but whatever.
So 63 x 96 -->
5 4 1 8
_ 2 7 _
_ 3 6 _
6 0 4 8
Why? Because 6x9 is 54, and 3x6 is 18 (first two digits and last two digits respectively), one diagonal gives you 27 (3x9) and the other gives you 36 (6x6).
Or if I don't have the luxury of seeing it on paper,
54/27 -> 567+36 = 603 -> 603/18 = 6048. (Remember that / represents skewed addition, meaning I add the two digits left and right of the slash and combine them into one, in this case I added 4/2 to get 6, and 3/1 to get 4.)
So how does this work? Well... let (ab)_10 and (cd)_10 be 2 digit numbers (a is the first digit, b is the second, and same for cd). Then, (ab)_10 = 10a + b and (cd)_10 = 10c+ d. So the two multiplied would be (10a+b)(10c+d) = 100ac + 10(ad+bc) + bd. Which is simple foil. But you can turn this "simple" formula into insanely quick mathematics, especially when you look for specific cases. But what did the first method mean? So get this in your head:
(ac goes here) (ad + bc goes here) (bd goes here) and let whatever carries they have move into the one to the left of it. So for example, if I was doing 92 x 94 (which I will show you a quicker trick in a bit),
92 x 94
(ac) (ad+bc) (bd)
8 1 / 4 / 8
_ 5
So it's 8648. The 5 is the "carry" from ad + bc since (9x4) + (2x9) = 54.
Now let's observe what happens when one of the digits are the same, as in this case.
(10a+b)(10a+d) = 100a^2 + 10(ad+ba) + bd = 100a^2 + 10a(b+d) + bd
The case where the second digits are the same is symmetrical.
So two quick examples:
62 x 69. 6^2 is 36, so the first two digits is 36...kind of. 2 + 9 = 11 x 6 = 66, so add 6 to 36 to get 42 and "tack on" a 6 to get 426. 2x9 is 18, so add 1 to 426 to get 427 and "tack on" the 8 to get 4278.
57 x 97. 5 x 9 is 45; added together is 14, 14 x 7 is 98, so add 9 to 45 to get 54 and "tack on" 8 to get 548, then 7^2 is 49, so add 4 to 548 to get 552, "tack on" 9 to get 5529.
(Note: What I'm doing here is adding one digit at a time instead of adding two. It's easier to add two one digits than to add one two digit. Hence "Divide and Conquer." So since I'm doing a skewed addition, adding 548/49 I'd add the 4 in the 49 first and then attack the 9 to the end.)
That's not very hard, is it? Wait! It gets crazier...
So the general formula for two digit squares is:
(10a+b)^2 = 100a^2 + 20ab + b^2. Which you can get pretty quick at in general, but there are some shortcuts. I personally, through the Vedic Mathematics systems, have a rule memorized for every ending digit. But I'll show you the cool ones.
Let b = 5. Then, 100a^2 + 100a + 25 = 100(a(a+1)) + 25. This is wonderful. This means you take the first digit, multiply by the thing above it (henceforth known as "upladder"), then tack on a 25 at the end. Examples: 65^2 = 4225 (because 6x7 = 42), and 95^2 = 9025 (because 9 x 10 = 90), and 999999995^2 = 999999990000000025. (Trick extension) This means any square ending in 5 can be done this way.
Let a = 5. Then, 2500 + 100b + b^2 = 25(100) + 100b + b^2 = 100(25 + b) + b^2. So you take the second digit, add it to 25, then tack on the second digit squared. Examples: 57^2 = 3249, 59^2 = 3481, 5401^2 = 29170801 (yes, I did do that in my head).
So 5 is a lovely number, as I'm sure you've deduced. The vedic mathematic system involves plugging in certain numbers to see what the results are. I will leave this up to you; the discovery will enable you to memorize the formulas better. For example,
(10a + 7)^2 = 100a^2 + 140a + 49 = 100a^2 + 100a + 40a + 40 + 9 = 100(a(a+1)) + 10(4(a+1)) + 9. So this is slightly complicated, but 87^2 you take 8 and you add 1 to get 9, multiply by 4 to get 36, 8 times it's upladder is 72, so you have 72/36 = 756 tack on 9 to get 7569.
Remember when I said "I'll show you an easier way to do 92 x 94?" Well, here it is. One of the gems of Vedic Mathematics, and the extension of this is so wonderful. So notice that 92 and 94 are really close to 100. Well, it just so happens that
(100 - a)(100 - b) = 10000 - 100(a+b) + ab
I'm sure you can see where this is going.
98 x 94. 98 has a "deficiency" (distance from the "close number," in this case 100) of 2, and 94 has a deficiency of 6. So add the two and you get 8. Subtract this from 100 to get 92, then 6 x 2 is 12, so the answer is 9212. Alternatively, you could just subtract one of the deficiencies from the other number to get the same first two digits. Verify that this is true.
But yeah anything near 100 I do with this. So 97^2? 9409. Remember that it's the DEFICIENCY and not the DIGIT; it's common to get this mixed up at first.
What's 93 x 95? 8835!
So happy maths lol. I hope this has encouraged some of you to put down your calculators and exercise your brain. Mental mathematics has been shown to be really good for you, and prevent things like Alzheimer's and things of that nature. And it keeps your brain moving, makes you sharp, and increases your active brain capacity. Or some nonsense like that.
But if you think that's it, it's not. I have way, way, way, way, way more. And I'll explain how I did 5401^2 in my head...and even how I did 7867654 x 568765 and missed only 3 digits lol. Happy mathing! I'm such a mathie.
================
Here's part 2, which I wrote today.
================
So last time we left off on 2 two digit numbers (man I hate saying that, especially, what if you wanted too to multiply two to two two digit numbers?), and today I'm going to extend upon that. I showed you the fives tricks, and the deficiencies trick, but today we're going to focus more on cross multiplying and extending those tricks.
A certain amount of memorization will have to take place (like really nice powers; anything that can speed things up you need), but don't worry about that.
1. The Problem
The problem with mental math isn't that you can't do it - anyone can do math in their head. The problem is that you can't remember the partial results. For example, when you're doing 8491 x 9525, by the time you've reached the second digit, you're forgotten the partial results. So the solution is to develop that minimizes the requirement for partial results as much as possible - and develop a way to help you remember the few partial results (which will turn into large amounts for large multiplications anyway). So with these two concepts, let us delve into the wonderful world of mental arithmetic.
2. Three Digit by Three Digit (and consequently, 3 digit by 2 digit)
Three digit by two digit is actually really easy; you just have to be able to remember things. You could repeat them to yourself again and again and again, but that's hard and we don't want things to be hard. Or you could invest some time in mnemonics - which I did, and it turns into something really useful. (Those who learned pegging should be really familiar with this).
But first, I'll need to show you the actual cross-multiplication pattern. As usual, we'll work left to right.
![[image loading]](http://i.imgur.com/fQqbdiv.jpg)
And this is wonderful, since you're just tacking on one thing at a time. The way I'm currently doing it is I have a running total, and I don't even add the numbers up, I just add each multiplication to it as I go. Which really isn't that bad.
Whenever you have a digit missing, just treat it as 0. If you have a 0, that's even better.
Special Case: Same Starting Digit
I showed you a trick for identical starting digit for two digits, but with three digits, it gets a bit more interesting, and, surprisingly, some cases are actually easier. So if you were to ask me what 124^2 would be, I would immediately reply 15376.
So this trick is based on the following:
(100 + b)(100 + a) = 10000 + 100(a+b) + ab
So like 103 x 109 = 11227, (1_12_27 because 3+9 = 12 and 3 x 9 = 27) and 106^2 = 11236 because 6+6=12 and 6x6 = 36. So 124^2 = 148/576 = 15376. I could do this for ages.
When you get into the 60's though, it gets kind of annoying. 169^2 = 1/138//4761 = 28561, but with practice, it becomes easy.
Changing the starting digits involves a mere exploration as to what happens:
(200+a)(200+b) = 40000 + 200(a+b) + ab. So now you have to multiply everything by 2.
212 x 215 = 45580 because 12 + 15 = 27 x 4 = 54, and 12 x 15 = 180, so 454/180 = 45580. You can extend this pattern, but once you cross fives, it gets annoying.
Case in point:
(700+a)(700+b) = 490000 + 700(a+b) + ab
723 x 743. So you'd add 23 and 43 to get 66, times 7 is 462, 23 x 43 = 989, so the answer is 49/462/989 = 5362/989 = 537189. With practice, this gets easier in your head. And take note of how few numbers you have to remember.
So here's how I did 5401^2 in the previous example. This one was particularly easy because 401 has a 0 and is so close to 400. So 401^2 = 160801, as I'm sure you can easily verify if you extended the method above with same starting digits. Remember the two digit beginning with 5 square? How you add the second digit to 25? Well, the trick can be extended. In this case, the number is 5401, so I'm going to add 401 to 2500 to get 2901. Why can I do this? (5000+a)^2 = 25000000 + 10000a + a^2 = 10000(a + 2500) + a^2. So now I just attach 160801 to the end of that... 2901//160801 = 29170801.
An alternate solution (for squares) is to break the numbers up in terms of 2 and 1, instead of 1 and 2. So deal with the first two digits and the last one separately. Let's take another look at 169^2. Here's how I would actually have done the problem.
So recall that (a+b)^2 = a^2 + 2ab + b^2. So let a = 10a to yield (the a's are not the same) (10a+b)^2 = 100a^2 + 20ab + b^2. So in this case, a = 16 and b = 9. I chose this because I know that 16 x 9 is 144, so that cuts down a lot of time dramatically. 20 x 144 = 2880, so we "just" have to add 25600 and 81 to that. So I'll use piecewise addition here (which means you only add one part at a time). 256 + 28 = 284, and 80 + 81 = 161. So 284/161 = 28561.
I use this technique in the 900's a lot. 900's are annoying, but I can do multiplication in the 90's almost instantly. Take 912 x 948 for instance. I mean this is kind of nice because the last 2 digits are both multiples of 12 and add up to 60, so if you did the other method (which would actually be faster in this case), you could do 81/540/576 = 864576. So I guess I should use another example since I already killed it using the old method lol.
So let's take 933 x 974. Ew, nothing nice about that. So (10a+ b)(10c+d) = 100ac + 10(ad+bc) + bd. This looks kind of daunting...but honestly, it's not that bad. So 93 x 97 = 9021 (remember, I multiplied deficiencies! But I guess I should cover last digits add up to 10; if it's same starting digit and the last two digits add up to 10, then (10a +b)(10a+c) = 100a^2 + 10a(b+c) + bc = 100a^2 + 100a + bc = 100(a(a+1)) + bc, so you multiply by upladder and take the last two digits and multiply them), but now the monster appears...ad + bc. This just looks ugly. But 93 x 4 = 372 and 97 x 3 = 291, and together is 372 + 300 - 9 = 663. So the answer is 902100 + 6630 + 12 = 9021//663/12 = 908742. Sometimes, there's just now way out of that, but imagine the other way...these take practice.
3. Four Digits and Above...
You just extend the cross multiplying pattern. Why? Because of distribution properties. I could use a sophisticated combinatorial argument to explain why this would work, but I'm lazy and I'll just draw a picture instead and you can prove this with algebraic induction if you want to.
So let's say we have a 7 digit x 7 digit (A MONSTER)
So we know that this number can have 13 to 14 digits, depending. So let's lay them out:
A B C D E F G
H J K L M N O
So, left to right:
1. A x H
2. AJ + BH
3. AK + BJ + CH
4. AL + BK + CJ + DH
5. AM + BL + CK + DJ + EH
6. AN + BM + CL + DK + EJ + FH
7. AO + BN + CM + DL + EK + FJ + GH
8. BO + CN + DM + EL + FK + GJ
9. CO + DN + EM + FL + GK
10. DO + EN + FM + GL
11. EO + FN + GM
12. FO + GM
13. G x O
And that's "all you have to do." Lol if you train yourself, this could take you like 30 seconds, if you could remember all of it...
So here's an example.
5501 x 3982. This one is a pain. Thank goodness for the fives.
So the layers are:
15,60,85,53,19,8,2-> 210/85/53/19,8,2 -> 2185/53/19,8,2 -> 21903/19,8,2 = 21904982. That wasn't THAT bad was it? Okay, I'll admit it, if it weren't for mnemonics, I woudn't be able to remember all these numbers. NO ONE CAN. Except maybe austistic savants, but none of us reading/writing this are autistic savants, I don't think.
4. Remembering the Numbers...How?
Chances are, the personing reading this did not know that every year, the United States hosts the International Memory Olympics. I bet even fewer people know that one of the events is longer number memorization - where they give you insanely long numbers and ask you to memorize them within like 5 minutes and recite them - without mistakes. I think even fewer of those people will know that these people will tell you, "I have an ordinary brain...just like everyone else."
Why? Because memory is a trained skill. The moment someone says "Oh, I wasn't born with a good enough brain to memorize things" I immediately facepalm and teach them link. If you want to learn link, google it, it's super simple, a few concepts will change your life. The Greeks did this.
Anyway, a particularly useful thing which I learned a while back was the mnemonic system for learning number memorization. Mnemonic is a thing that helps you memorize/remember something.
So if I asked you to remember tourniquet or 14271, which one will your remember better? Equal chances for either one. What about 24897623490682769028620 vs "Nora VIP, kishin' more posse Shiva knock ships knife shins." Chances are, they're both a bit hard to remember, but with a bit of link, the second one is probably much easier to remember. (Lol Shiva). Here's the catch. To me, they mean the exact same thing. And it wasn't hard for me to remember the second one. At all. That's how I memorized the number, which was like 23 digits or something.
This is like the peg system; for example, if I wanted to remember the 63rd item on a certain list, I'd look for Jean Park (whom we called Shim) because 63 corresponds to Shim, or chime, whichever one I'm feeling at the moment. I should develop consistent pegs but I'm lazy...
So each consonant sound corresponds to a number. I should probably get better at making sense out of these words that I come up with, but I'm still kind of noob at this, and I'm working on it.
0 - s, soft c
1 - t, d
2 - n
3 - m
4 - r
5 - l
6 - sh, hard ch, j, soft g
7 - hard g, k, hard ch, hard c
8 - f, v
9 - p, b
And you can see that my example, albeit a horrendous one, does really mean the exact same thing. So I still remember the result to that one problem from the other note (gnome be moving) which is 23938.
So I was learning how to do this, and I thought...why not do mental arithmetic with this trick? So what happens is I start adding letters to letters - but I don't think of it as such, I think of it as changing. And this is so weird and takes practice but it's ridiculously efficient.
There are two ways to do this - but I like to use two numbers per word, and change it if I have to. So let's do that one problem from the other note.
7867654 x 568765 (Kiefer shock sheller x Lush pack shell)
Thank goodness 568765 has a missing digit. So the first row is blank, and the 8 multiplies a blank digit in the next one... so we start with 7 x 5 which is 35.
Then 3 by 2...so 7 x 6, 8 x 5, and 6 x empty. So really just 7x6 and 8x5 which is 82. So we're now at 432. I'm confident that the first digit will not change with the following results, so I now store the first few digits... Ramen (432). The next layer is 7x8, 8x6 6x5, and 7x0, which yields 134. So m + t = r and n + m = l, so rarel4 = 4454, or rereeler. Anything to remember it. Typically I don't even care about the meaning of the word; I just remember the sound of the word and that's usually enough. So the next layer is 7x7, 8x8, 6x6, 7x5 (which I swear is a coincidence), which yields 184 (I used a pythagorean triple there LOL) so rereeler -> rereesher -(add 1 to second to last digit)> rereeken -(add 1 to second to last digit, change last digit to 2)> rereckon -(tack on 4)-> rereckoner. (Note: I'm doing it layer by layer because I'm too lazy to type but normally I add one bit of each layer at a time). Now...the next layer...7x6, 8x7, 6x8, 7x6, 6x5 which is 218. Rereckoner...(add 2 to second to last digit).. rereckorer ..(add 1 to last digit)...rerechoral...(tack on 8) rerechoralf. The next layer! 7x5, 8x6, 6x7, 7x8, 6x6, 5x5 = 242. So rerechoralf...rerechorakoff...rerechorafiff, rerechorafinninn. (For the record, we're at 4474822, and if you check at this point, you'll see the first few digits are correct)
So we're halfway done lol. That would've taken me around a minute...I'm trying to cut it down to 2-3 seconds, but it's hard lol.
Continuing on, the next layer is 8x5,6x6,7x7,6x8,5x6,4x5 which is 223 I think. So rerechorafinninn -> rerechorafirir->rerechoraforirum. At this point I'm going to start a new word - rerecorafor, and then rum. So we'll remember rerecorafor, and work with rum. The next layer is 6x5,7x6,6x7,5x8,4x6, which is 178. So rum -> lame -(I see 3+7 is going to carry)> shame -> Jesse -(tack on 8)> Joseph. The next layer is 7x5,6x6,5x7,4x8, which is 138. So Jotef, Joneff, Jonet-> Janet, tack on 8 to get Janitoff. The next layer is 6x5,5x6, 4x7 = 88, so Janitoff -> Janinoff -> Janinosh -> Janinoshiff. The second to last layer is 5x5,6x4, which is 49, and the final layer is just 20, so we'll just add 510 to Janinoshiff, which is Janinokiff -> Janinokim -> Janinokimits.
So we have rerecorafor Janinokimits. As dumb as that sounds. Which is 4474846227310. Which is right.
So yeah...don't worry this'll happen way faster in your head than it took me to type out ^^
I'll post part 3 later. Yes, there's more.
=======================================
Part 3.
Yes, believe it or not, there is actually more madness. A lot more madness.
So we've been through the insane multiplications, as well as the sick memory systems, and we're going to take a step back and discuss some other tricks that just might help you here and there.
1. The Square Identities
Remember the square identities you learned in Algebra?
(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2 - 2ab + b^2
Well, we've used both of those in determining squares. But there's another one.
(a+b)(a-b) = a^2 - b^2
Which actually has a plethora of uses in terms of the maths.
For example, say I want to multiply 509 x 515. Which wouldn't be that bad if you were to do it the normal way; you'd get 25 as the starting digits, 15+9 = 24 x 5 = 120, and 15 x 9 = 135, so the answer would be 25/120/135 = 262135. But there's a faster way you could have done this.
Notice how close the two of those numbers are. We use this technique (this is found in both Vedic mathematics and Arthur Benjamin's book, but he does the reverse of this) to do numbers close to each other, or use numbers close to each other to find squares.
So to do this example, notice that 509 and 515 are 6 apart. So halfway between them is a distance of 3 from either one, which is 512. How lovely, a power of 2! I'll get into higher powers later, but 512^2 is easy...25/12,144 = 262144...now we just need to subtract 3^2 (because that was our b) to get 262135. Which one is quicker?
That's up to you, but personally if the numbers get really close together, I'd use the second one. The first method wasn't that bad with this specific example because it began with 5.
What if you were asked to multiply 739 x 749? That's when you groan. So our distance is 5 and the midpoint is 744. So we just have to square 744 and subtract 5 from that...so let's do that! 44^2 is the same as 11^2 x 16 = 11^2 x 8 x 2, which is 1936 (that's one way to do it, another would be 16/32/16 = 1936) and we instantly know that 7^2 is 49, so we'll save those. 1936 is tip mash, so I see myself mashing the tip into the waiter's face. As mean as that is
Now we multiply 7 x 88, which is 560 + 56 = 616. So the answer is 49/616//1936 [remember in 3 digit numbers, each of the positions are only shifted by 2], which yields 553536. So we just subtract 25 from that to get 553511. It might seem longer, but that's because I typed an anecdote in there lol.Let's do it the normal way. 739 x 749. Well, the initial digits are 49, so we'll set that aside. 39+49 = 88 x 7 = 616. 39x49...3x4 = 12, 3+4 = 7 x 9 = 63, 9x9 = 81, so that then becomes 12/63/81 = 183/81 = 1911. So 1911 is the name of a gun which makes it really easy to remember so I just imagine myself holding one. So then 49/616//1911 = 5516//1911 = 553511. The first one was less thinking honestly.
You can use the reverse of this to find small squares too. 74^2 - 4^2 = (74-4)(74+4) so 74^2 = (70)(78) + 4^2 which is the method Arthur Benjamin uses. Personally I suck at divide and conquer so I don't use it; I find its reverse to be more useful. [The answer is 5476 by the way]
2. Peeling
Most math teachers are very familiar with this technique and use it. Unless they're incompetent. [Lol jk]
But what peeling is is recognizing factors that you can use, and manipulating these factors to make multiplications much, much easier.
For example, 576 x 121. I recognize immediately that both of these are squares, of 24 and 11 respectively. 24 x 11 is a nobrainer, 264. Now I just have to square that, which honestly isn't so bad; initial digit is 4, then 4x64 = 256, and 64^2 = 4096 [I remember most of my poewrs of 2] so it's 4/256 = 656 -> 656//4096 = 69696.
That other example I did...involved 48 x 12. Well, I saw immediately that 12 x 4 = 48, so I could "peel" one of the 2's off of the 48 and give it to the 12 to turn it into a square, which is 48 x 12 = 24 x 2 x 12 = 24 x 24 = 576, since I had it memorized.
You should start to have an intuitive sense of the numbers under 100; what are factors of what? What's prime, what's not prime, and what's divisible by what?
57 = 19 x 3. Because 57 is 3 less than 60 and 20 x 3 = 60. 91 = 7 x 13 because 91 is 21 greater than 70. If you observe factors until 100 for a while, you'll start to get an intuitive sense of them. 85 = 17 x 5, and 15 x 7 = 105, etc, etc..
So some other examples...
I'm really good at halving/doubling things. So whenever I have a power of two, I sometimes get lazy and just do this.
232 x 128. So all I have to do is double 232 7 times... 232 -> 464 -> 8/12,8 = 928 -> 184/16 = 1856 -> 36/102 = 3702 -> 7404 ->14808 -> 28/1616 -> 29616.
One thing to look for is inherent factors, or if they are a certain multiple apart, preferably a perfect square. Why?
Well, take 35 x 140. Well, this problem is really easy - it's 4900. Because 35 x 4 = 140, and 4 = 2 x 2, you can "peel" one of the 2's off of the 140 and give it to the 35. So 35 x 2 x 2 x 35 = 70 x 70 - which any of us can do almost instantly.
144 x 16 is a little less obvious; that requires knowing that 144 = 16 x 9. So I give one of the 3's in the 144 to the 16 to get 48^2 = 2304. So 144 x 16 = 2304.
3. Cubes (and higher powers of 2 digit numbers)
For whatever reason, we're going to now start cubing numbers in our head [all this was just me wondering to myself, how far can I push myself?], probably because Poli Sci is just so boring...
So first, memorize the cubes 1-10. This wasn't in the prerequesites because - well - you didn't really need them to get started, but please do indeed memorize these.
1 - 1 [obviously [we have a rapist in Lincoln Park [I'm super tired as I'm typing this]]]
2 - 8
3 - 27
4 - 64
5 - 125
6 - 216
7 - 343
8 - 512
9 - 729
10 - 1000
Look how quick these things blow up!
*Just a side note, 11^n is really interesting, because it's kind of based on how well you know your Pascal's triangle. 11^3 = 1331 [the first coefficients are the actual numbers because they add the same way as 11 multiplies], and 11^4 = 14641. 11^6 = 1,6/15/20/15,6,1 = 1771561. You can memorize the left half of Pascal's triangle with mnemonics [or other formulas] and then use symmetry to recall the other to do these, and Pascal's triangle will come up with higher powers.
So two digit cubes are not that bad - the generic way [and the only real thing you can do] is binomial theorem.
So recall that (a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3. This is how we'll deal with cubes for the most part.
(10a+b)^3 = 1000a^3 + 300a^2 b + 30ab^2 + b^3 = 1000a^3 + 30ab(10a+b) + b^3
Personally, I like grouping the middle terms.
So the patterm goes
_ _ _ _
1 2 3 4
1. a^3
2. -
3. 3ab(10a+b) [carry all but last 2]
4. b^3 [carry all but last]
So this actually doesn't look ALL that bad. Let's try a number like 24. Keeping it low because these have a tendency to go insane. So the initial digit is 8. So we'll reverse that. The middle term is 24 x 2 x 4 x 3 = 576. The last term is 64. Now it's really important that you know where everything goes. The formulas tell us that. So since the only thing I dropped was the 10, the 576 will just be moved over one place. So our answer is 8/576/64 = 1376/64 = 13824. You COULD square 24 and multiply 2 to that, but this is slightly faster.
Let's try an insane one. Like 98^3. Well, I'd probably use square trinomials here... (a-b)^3 = (a-b)(a^2 - 2ab + b^2). So what this means is that if a = 100 and b = 2, then this problem is still almost as hard...but not quite. a^2 + b^2 is easy, that's 10004. 2ab is 400, so that leaves us with 9604. So now the problem is 9604 x 98...which isn't THAT bad. 9604 x 100 = 960400 -> 960400 - 2(9604) = 960400 - 19208 = 941192. Whew we used a really roundabout solution...
There really is no easy way to do this. The formula is about as simple as it gets...
So let's do this using binomial theorem. 98^3. The first term is 729 -> Kimbo [Kimbo Slice, that's hard to forget]. Next is 9x8x98 x 3. So 9x8 = 72 x 98 = 7200 - 144 = 7056 x 3 = 21168 -> knitted shave [picture myself knitting my stubble]. Finally, 8^3 is 512 -> lighten. Do you see why we memorized all of them? Kimbo + knitted [since shave is the noncarry digits] so kimbo -> bimbo -> Binby -> Berries. Then shave + light [the en is the last digit] -> ohnoes, a carry! [1] tave -> tape. So berries -> parrot, so the answer is parrot tape inn = 941192.
The second way is almost lightning if you are fluent with the mnenomics [which I am not, so I would do it the first way, but I notice I spend most of my time struggling to decipher the mnemonics lol that's how it always is, and adding with them is weird too.]
We'll briefly examine the 5 cases to see if they're of any use to us, then move on to fourth powers and things with three or more digits.
(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3
Let a = 50. then (50+b)^3 = 1000(125) + 3(50^2)(b) + 150(b^2) + b^3 = 125000 + 7500b + 150b^2 + b^3 = 120000 + 5000 + 7000b + 500b + 100b^2 + 50b^2 + b^3 = 120000 + 1000(5+7b) + 100(b^2 + 5b) + 50b^2 + b^3
Not gonna lie, I'm not going to bother memorizing that. By symmetry, plugging in b would yield a similar result, except you'd know the last digits. Which is kind of pointless because you'd know them anyway.
But it's the exploring that matters, I guess.
Higher powers are a nightmare...if I wanted to do a 4th power, I would just square the 2 digit number, then square the resulting 3-4 digit number. But let's work out the formula anyway.
(a+b)^4 = a^4 + 4a^3 b + 6a^2 b^2 + 4ab^3 + b^4
(10a+b)^4 = 10000a^4 + 4000a^3 b + 600a^2 b^2 + 40ab^3 + b^4
There's no real need memorizing fourth powers of single digits - they're so easy to figure out. But here's a list.
1 - 1 [no sarcastic comments here]
2 - 16
3 - 81
4 - 256
5 - 625
6 - 1296
7 - 2401
8 - 4096
9 - 6561
10 - 10000
These things blow up RIDICULOUSLY fast.
(10a+b)^4 = 10000a^4 + 4000a^3 b + 600a^2 b^2 + 40ab^3 + b^4
= 10000a^4 + 40ab(100a^2 + b^2) + 600a^2 b^2 + b^4.
Or you could say 10000a^4 + 10ab( 4(100a^2+b^2) + 60ab) + b^4
_ _ _ _ _
1 2 3 4 5
1. a^4
2. blank
3. 6a^2 b^2
4. 4ab(100a^2+b^2)
5. b^4
For the second formula,
1. a^4
2. -
3. -
4. ab(4(100a^2+b^2) + 60ab)
5. b^4
I see a trick I might be able to pull with the middle terms - if I factored out the ab from the middle terms ab(4000a^2 + 600ab + 40b^2) that looks so close to a perfect square trinomial; I will explore this algebraically and see if I can add/subtract things to turn it into one that might be nicer. But until then, stick to this, or square something twice. I've never done this, but here's how I would derive methods to do so, so let's try one.
55^4. Now, if I square then square...55^2 = 3025. 3025^2 = 9150625. Which wasn't very hard because I had the foresight that 55^2 had a zero in it [especially in the second position] so squaring it would be a cinch.
Now let's try something ugly with the new formula...let's try 86. Nothing nice about that...so 8^4 is 4096 -> rice push. Then 6a^2b^2 just looks painful. They all do...6^2 = 36 x 64 = 18/48/24 = 2304, 2304 x 6 ->12304 -> 13804 -> 13824 -> tame finer -> Timiff nor [because we have to add Timiff to rice push] rice push -> rite push -> Rhine nush -> Rhine mull. So our partial result is Rhino mall nor. Then 4ab(100a^2 + b^2)...4(48)(6436) = 1235712 [oh goodness] and we have to add all the digits except the last one...so we'll add these two at a time. Rhino + 12 = Lore. Mall + 35 = kiss. Nor + 71 = pill. So lore kiss pill new. [the new is the 2 at the end]. So finally, b^4 = 1296, so we see that only pill and new get added to...but it looks like a carry is coming so let's bring kiss into the equation. so pill + 12 = [1]sick so change kiss -> kit and keep that. new + 9 = 1[tea], so change sick to sofa. Tack a 6 on the end of tea to get tush, so the answer is lore kit sofa tush = 54710816. I'm off by 10000 [the answer is 54700816] but I'm too lazy to look for it...
4. Calculating the Day of the Week
It's actually really easy. Some people use anchor dates and those systems but there are easier ways.
But first I must introduce arithmetic modulo 7. So in our case, all it means is that you care ONLY about the remainder when you divide by 7.
So the equation goes:
p = m + d + c + y + floor(y/4) (mod 7)
p is the day of the month [0 is Sunday, 6 is Saturday, 3 is Wednesday, etc]
m is the "month number," I'll provide a chart
d is the day.
c is the "century number," I'll provide a chart
y is the last two digits of the year.
floor(y/4) means you take the last 2 digits of the year, divide by 4, and round down.
Add all these up and divide by 7, and the remainder is the day of the week.
Century Number [note: I will not cover Julian calendar; so don't do any dates before 1752. If you want to go before 1752, go here: http://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week this is also where I learned this method; I was using a more inefficient one until I found this one]:
If you divide by 4 and the remainder is 0 [the number is congruent to 0 mod 4], then c = 6
If remainder is 1, then c = 4
If remainder is 2, then c = 2
If remainder is 3, then c = 0 [which is really nice for 1900's]
Month Number [note, if the last two digits are divisible by 4, it's a leap year]
Jan: 0 (leap year -1)
Feb: 3 (leap year 2)
Mar: 3
April: 6
May: 1
June: 4
July: 6
August: 2
Sep: 5
October: 0
November: 2
December: 5
So take the day the Declaration of Independence was signed, 7/4/1776. So I like starting with the months and day. And the nice thing about this is that you don't need to remember any partial results. So July means m = 6. d = 4, add them to get 10, but we're in mod 7 so it's really 3. Look at the year, it's 1700, so that means c = 4. So 4 + 3 = 7, so we're back at 0. So we only need to look at the last 2 digits of the year now. 76 itself is 6 mod 7, so we remember 6 and examine 76/4 = 19. 19 + 6 = 25 , which is congruent to 4 mod 7, so it was a Thursday.
It's honestly not very hard, just memorize the numbers and you're fine.
And believe it or not...there's actually more. I thought I'd be able to cover everything, but there's so much more...
=======================
Part 4
=======================
This will most likely be the last installment unless I think of something else to write about [like division or subtraction, but that stuff isn't as sought after as multiplication; and frankly I just don't care about them that much but I might look into it later], and today's is probably going to be the craziest.
So this is a stage that I have not yet achieved yet, but it is something that I'm going to strive toward.
So here is where you should aim once you're really, really fluid with the whole concept of "addition/multiplication of letters." At this point, we should be able to take up to 3 digits at a time and add it to another 3 digit word, and have this happen insanely quickly. I'm not quite there yet, but I have developed methods of where I would like to get, and hopefully I get here someday. [The only thing I really need to do is get good at the phonetic system; once the phonetic system is mastered, all this stuff becomes easy, and the only thing that changes is the time element, because words are easy enough to remember, right?]
On a side note, I developed the phonetic system partially because I wanted to compete in the memory olympics using them. But upon closer examination of the rules...YOU'RE ALLOWED TO ENTER EACH DIGIT AS THEY COME. That's too easy! You just use cross multiplication and work your way right to left and that's hilariously easy. I want to go through two twenty digit numbers [the current world record] and not be allowed to write down a single result until I'm done. That's the goal at the end of the day here. I want to set the world record for that.
Anyway, this time, we extend our results to work with more than two numbers.
1. Revisiting Addition of Multiple Numbers
We dealt with addition of many numbers, but last time, we were able to see them in front of us. This time, we're not. [For the Memoriad competitions, there's the Flash Anzan where they give you 0.3 seconds to look at each number, which is insane; I think this system will never catch up to Sudoban mathematics in terms of that, but hey, I can try, right? XD] Not only is this insanely useful in just about every multiplication I described - this becomes pivotal in doing the multiplications I described, and becomes a very useful skill to have anyway.
So we have now learned how to take numbers and turn them into words, and vice versa. I encourage you to practice the two until they become nearly synonymous. This is what I am currently working on now. At the same time, start doing "word addition problems." Which means you begin with two words (or more), and then you add them, and end up with another word, and then turn that word into a number. Or you could start with two numbers, turn them into words, and then do the addition, and spit back a number to see if you got it right. This becomes crucial.
So pee + ma = tin (9 + 3 = 12), wine + beer = push (02 + 94 = 96)
So now, we simply memorize letter combinations that add up to 10 do the same "add up to 10" trick we did, but this time, it's with a bunch of numbers that we memorize as they appear to us.
So let's say someone calls out 23025 + 63953 + 52190 + 48965 + 54321
But we don't have the luxury of the numbers in front of us - the only thing we have in our head is Nima's nail (I know a guy named Nima), chimp lime, lint bees, rave bushel, and liar mint. Now it gets a bit tricky, since I need to use pictures to remember these things. So I literally see Nima holding up a bloody nail [blood scars me and makes it easier to remember], a monkey that climbs a lime tree to get limes and eats it and turns into a lime, a rave bushel is a unit of measurement for how much "beverage" one consumes at a rave [I don't know, but it's easy to remember] and it's easy enough to come up with a picture of that, lint bees - I see bees harvesting the lint out of someone's belly button - and surely that's scarring enough. Finally, I see myself being turned into a mint - and eating mints after having my mouth washed out with soap for lying. So now I have all of those words in front of me. This needs to happen at light speed.
You can use link if you need to to string them together - after all, you CAN add them one at a time. I'm doing this in the ballroom of my memory palace, where I have a series of pedastals set up that can accomodate up to 20 things [the room is specifically reserved for mathematics], and it's easy enough to see where I put everything. So I seem to keep everything there.
So I have a look around and see my five numbers. Now let's add them quickly.
Nima's nail
CHimp lime
Lint bees
Rave bushel
Liar mint
Well, the two l's make a 10, the chimp and rave make a 10, and nima is left over. So we have 22 so far - nun. Looking over, raVe and liNt, as well as liaR, niMa, and chiMp make 10, a lucky concidence - so that's 20. Nun becomes Norris. [Like Chuck Norris]. Then, nima'S, chimP, linT, Bushel, and Mint - the S doesn't do anything, P and T make 1, and B and M make 12, so we're at 22, so Norris becomes Nornin. The next row, n,l,b,sh,n means b and n make an 11, and the rest make 13, so we're at 24, so nornit becomes norn roar. The final row, l, m, s, l, t makes 14, so norn roller. Which is 242454.
This needs to become as fluent as the back of your hand. Then we can go on to the insane stuff.
2. High Powers of Low Numbers
This trick is pretty simple - use the power rules. For example, in my Number Theory class - the teacher put the example on the board, "2^20 (mod 41)," well, no one wants to tell me what 2^20 is, do they? And I immediately yelled out "1,048,576!"
But that wasn't very difficult. I knew that 2^20 = (2^5)^4 = 32^4 = 1024^2 = 1048576.
Try to break things up so that they're relatively simple; try to look for cases where some multiplications may be easier than others. For example, when doing powers of five, always try to keep the first digit a 5 so you can use that trick again and again...with powers of 2 (and consequently 4), try to keep it around 1024, since that's just such a nice number to work with; it starts with 1, there's a 0, and the last two digits are sub 33. What else could you hope for?
So an example: 5^12. So I could do this in 125's and break even or 625's... or even 15625's (which is 5^6 power, obtained from squaring 125). Multiplying by 5 again wouldn't end in 5, so I'm doing to stop here and square this. As painful as that sounds. But I chose this because the second digit is a 5, and the first digit is a 1, so when I break it up to (10000+5625)^2, I have two rules I can use - the rule with the 1 as leading digit [which is so nice] and the squares beginning with 5 rule. So if we square this number, the leading digit is 1, 5625 x 2 = 11250 and 1/11250 = 21250 -> nut nails. Then 5625^2... 25/625 = 3125 -> mad nail. So I have two kinds of nails - nut nails and mad nail. Anyway, 625^2 is pretty easy - 62x63,25 = 390625. So mad nail//390625 -> dan shore 0625 = mad shore such nail -> mad shores Jenelle. (31640625) [grouping them in 3's is easier for calling out the numbers] So I go back to my toolbox and get my nut nails. I'm expecting a 9 digit number out of this one.
So it's going to be nut nails //// mad shores Jenelle. So we're going to add those and get no roar tours, Jenelle for the answer, which is 244140625.
3. The One Thing I See Missing in Mental Math Competitions
The challenge that I neer see in mental arithmetic competitions is that of multiplying many numbers together. I mean it's easy to see that you can take partial results and multiply them together, but that's the same as doing like 3 little multiplication problems in one, which there's nothing wrong with...but there's a "faster" way. Faster depending on how you like to look at it.
So I showed you the general formula for multiplications of two numbers - you would just cross multiply all the way through. So first I'm going to show you how to do many two digit numbers, and then I will show you how to do many numbers of as many digits as you like.
So first let's try 3 two-digit numbers.
(10a+b)(10c+d)(10e+f) = 1000ace + 100(ade+bce+acf)+10(bde+adf+bcf)+bdf.
...what? Well, once you see the pattern, it's actually not that bad.
A B
C D
E F
So ace is obvious; it's the first row! Same with bdf. But "ade, bce, and acf" get weird. But don't worry; it's just all the different ways you can take two things from the first column, and one thing from the second. And "bde, adf, and bcf" are all the different ways you can take one thing from the first column, and two from the second! And this pattern extends infinitely! So theoretically, if you could remember everything [and had enough time], you could do an 81 eighty-one digit number multiplication! I mean, you're going to be there for a really, really, realy long time but still.
Let's do an example:
91
84
72
(you should practice them looking at them first; these will take a while to do without looking, not to mention the mnemonics need to be REALLY solid)
Well, the first row is 9 x 8 x 7; always try to multiply by 9, so 56 x 9 = 504 = laser. Then, take the first 2 from the first column, and the last from the second column, or 9 x 8 x 2 = 144 = terror. Then 1-2-1 [the previous was 1-1-2] which is 9 x 4 x 7 = 252, and terror + 252 = my Peach [like the princess]. Now, 2-1-1, which is 1 x 8 x 7 = 56, and my Peach + 56 = rollin'. So since this only has a one digit spacing, we need to add the first 2 digits to terror, which is laser//rollin' to yield lair pin. Now we do 1-2-2 (72), 2-1-2 (16) and 2-2-1 (28), and we add that up to get 116. So we add 11 to pin and reserve the 6. 11 + pin = sim tack on 6 to yield sammich. But there was a carry there, so lair -> lol, so lol sammich, and the final digit is easy, it's 8, so lol sammich fffuuuuuu or something idk. Which is 550368.
Extending this [I will not do examples; they take forever LOL], basically just go through each row and find a way to make all the "column sums" the same. So now, instead of the first row being labeled "1," it must be the highest number. So I'll give you an example.
3 2 1
A B C
D E F
G H I
So to multiply this, recognize that the first term has 6 zeros behind it (because that is how many non-leading digits there are; ADG are the leading digits and the rest are BCEFHI), and each term is just shifted over one to the right. So we need to draw 7 blanks.
1. 3-3-3 (9), which is ADG
2. Row values should sum to 8, so 3-3-2, 3-2-3, 2-3-3, or ADH + AEG + BDG
3. Row values should sum to 7, so 3-3-1, 3-1-3, 1-3-3, 3-2-2, 2-3-2, 2-2-3, or ADI + AFG + CDG + AEH + BDH + BEG
4. Row values should sum to 6, so 3-2-1, 3-1-2, 2-1-3, 2-3-1, 2-2-2, 1-2-3, 1-3-2 or AEI + AFH + BFG + BDI + BEH + CEG + CDH
5. Row values should sum to 5, so 3-1-1, 1-3-1, 1-1-3, 2-2-1, 2-1-2, 1-2-2, (you get the point)
6. Row values should sum to 4, so 2-1-1, 1-2-1, 1-1-2
7. Row values should sum to 3, so 1-1-1
Amount of products per row: (1-1, 2-3, 3-6, 4-7, 5-6, 6-3, 7-1) or 1 3 6 7 6 3 1. This can be confirmed by letting everything equal 1, and you will get 111^3 = 1367631. If you study combinatorics, picking which ones will add up to what will be quite simple.
One last example, and I'll only write out the product rows, and not the letters.
2 1
A B
C D
E F
G H
I J
K L
M N
Seven two digit numbers. I can see that there are going to be 8 rows.
1. Row numbers add up to 14. That's easy...2-2-2-2-2-2-2
2. Row numbers add up to 13. 1-2-2-2-2-2-2, 2-1-2-2-2-2-2, ... , 2-2-2-2-2-1-2, 2-2-2-2-2-2-1
3. Row numbers add up to 12. This is where it gets really impractical, because you can see how many different ways there would be to add seven 1's and 2's to get 12 (6+5+4+3+2+1 = 21)...so for this case, I would multiply them in pairs, and then again in pairs to get the final answer.
So this is kind of a pain, and I have yet to explore this, but it does seem quite do-able in the head for not too many numbers. Just make sure you don't miss anything.
4. Big Numbers by Small Numbers
We will end on a not-so-impressive note - big numbers by small numbers. The Trachtenberg system is extremely potent at this - but the only reason I don't use it is because it goes right to left, as opposed to left to right.
So this is only a modified version of cross-multiplication, except we pick however many digits the small number has, and we just do that many for the cross multiplication, and then stick the first digits multiplied together and the last digits multiplied together on it. Or you can do them in order; both are equally good. Here's what I mean.
Take 346905 x 54.
So you'd take 3x5, then 34 [cross] 54 = 32, then 46 [cross] 54, then 69 [cross] 54, and so on until you get to 05[cross]54. Then you'd skew add 5x4, and you'd get an answer. I'm too tired to do it.
So I hope you learned something from reading this lol.
Hope you didn't fall asleep. Mental arithmetic olympics, here I come!
=======================
Addendum
Danglars provides a way to remember the mnemonic system
+ Show Spoiler +
On January 31 2014 19:53 Danglars wrote:
Ever since reading (being read to actually) Cheaper by the Dozen when I was a kid, I loved the 2digit multiplication tricks:
37 times 43, find the number halfway between, and add/subtract 3: (40-3) x (40+3), which equals 40^2 - 3^2 or 1591.
Memorizing the squares up to 30 is crucial for in-head math.
It builds mental toughness.
Impress kids (I sometimes tutor in my free time). Mentally check their answers before they've even stacked the numbers.
You WILL be able to multiply two-digit numbers faster than somebody with a calculator can type them in. You WILL be able to get exact numbers on discounts and taxes to the amazement of your friends (Dress 34.99$ but 15% off? You have its $29.74 cost in 5 secs (You will use 35$ & 0.0085 in process). You have its cost with sales tax of 32.12$ in another 5 seconds or less.
I kid you not, if you learn 2digit multiplication, you will save time on tests, checking your work, and fly past those with a calculator on the basics.
With three digit numbers, I generally already have the calculator out for other reasons, so I generally stick with 2x2, 3x2, and various mental tricks if I'm okay settling for a 3digit approximation +/- 1% accuracy.
It deserves it's own topic! The modern mneumonic system is a thing of beauty. Important constants memorized in seconds. The way I learned
0 - s, c, soft z - z for zero, or c for cero
1 - t, d, - one downstroke t d
2 - n has 2 downstrokes
3 - m has 3 downstrokes
4 - r is the fourth letter of four
5 - L is roman numberal for 50
6 - cursive j is similar to 6
7 - build a k with 2 7's
8 - f, v - cursive f looks like an 8
9 - p, b - flip a capital p to get a 9
WHY - unused
Square root of 3 is 1.732 - Picture a kimono of 3's. Kimono KMN 732. And we already know the 1.
Ever since reading (being read to actually) Cheaper by the Dozen when I was a kid, I loved the 2digit multiplication tricks:
37 times 43, find the number halfway between, and add/subtract 3: (40-3) x (40+3), which equals 40^2 - 3^2 or 1591.
Memorizing the squares up to 30 is crucial for in-head math.
It builds mental toughness.
Impress kids (I sometimes tutor in my free time). Mentally check their answers before they've even stacked the numbers.
You WILL be able to multiply two-digit numbers faster than somebody with a calculator can type them in. You WILL be able to get exact numbers on discounts and taxes to the amazement of your friends (Dress 34.99$ but 15% off? You have its $29.74 cost in 5 secs (You will use 35$ & 0.0085 in process). You have its cost with sales tax of 32.12$ in another 5 seconds or less.
I kid you not, if you learn 2digit multiplication, you will save time on tests, checking your work, and fly past those with a calculator on the basics.
With three digit numbers, I generally already have the calculator out for other reasons, so I generally stick with 2x2, 3x2, and various mental tricks if I'm okay settling for a 3digit approximation +/- 1% accuracy.
It deserves it's own topic! The modern mneumonic system is a thing of beauty. Important constants memorized in seconds. The way I learned
0 - s, c, soft z - z for zero, or c for cero
1 - t, d, - one downstroke t d
2 - n has 2 downstrokes
3 - m has 3 downstrokes
4 - r is the fourth letter of four
5 - L is roman numberal for 50
6 - cursive j is similar to 6
7 - build a k with 2 7's
8 - f, v - cursive f looks like an 8
9 - p, b - flip a capital p to get a 9
WHY - unused
Square root of 3 is 1.732 - Picture a kimono of 3's. Kimono KMN 732. And we already know the 1.
Gowerly provides a way to quickly multiply numbers that end in 5.
+ Show Spoiler +
On January 31 2014 21:24 Gowerly wrote:
I've been trying a bit of this.
I'm not at that level, but I have a couple of tricks for "common" multiplications.
Numbers ending in 5 is good and easy (assuming they're reasonably close)
Squares of numbers ending in 5 are the easiest.
5 x 5 = 25
15 x 15 = 225
25 x 25 = 625
...
55 x 55 = 3025
...
95 x 95 = 9025
The way these work is as follows:
ignore the 5s on the end. That just becomes 25.
take the other digits and add one to one of them, multiply them together and stick them on the front.
for 95 x 95:
5s become 25
9s become 9 x 10
9025.
If you've memorised all the squares to 30 (and can add one more to them) you should be able to square any 3 digit number ending in 5 in under a couple of seconds.
For other numbers ending in 5 it's slightly more tricky.
I've found the general formula is thus.
- Ignore the 5s again.
- Add 1 to the larger of the two numbers
- Multiply them together, stick on front of the 25 as before
- Add 50 * difference between the numbers without the 5.
Easy Example:
225 x 255
This becomes 22 x 26 (can do 24 ^ 2 - 4) = 576 - 4 = 572
becomes 57225
Then add 150 ((25 - 22) x 50)
57375
I've been trying a bit of this.
I'm not at that level, but I have a couple of tricks for "common" multiplications.
Numbers ending in 5 is good and easy (assuming they're reasonably close)
Squares of numbers ending in 5 are the easiest.
5 x 5 = 25
15 x 15 = 225
25 x 25 = 625
...
55 x 55 = 3025
...
95 x 95 = 9025
The way these work is as follows:
ignore the 5s on the end. That just becomes 25.
take the other digits and add one to one of them, multiply them together and stick them on the front.
for 95 x 95:
5s become 25
9s become 9 x 10
9025.
If you've memorised all the squares to 30 (and can add one more to them) you should be able to square any 3 digit number ending in 5 in under a couple of seconds.
For other numbers ending in 5 it's slightly more tricky.
I've found the general formula is thus.
- Ignore the 5s again.
- Add 1 to the larger of the two numbers
- Multiply them together, stick on front of the 25 as before
- Add 50 * difference between the numbers without the 5.
Easy Example:
225 x 255
This becomes 22 x 26 (can do 24 ^ 2 - 4) = 576 - 4 = 572
becomes 57225
Then add 150 ((25 - 22) x 50)
57375
arbiter_md provides a trick for the number 142857
+ Show Spoiler +
On February 01 2014 00:37 arbiter_md wrote:
Related to this topic, I like the trick with number 142857, which you obtain from dividing 1 / 7.
So, besides 142857 * 7 = 999999, you also need to know the following feature: Multiplying by x < 7, the result will be the same number rotated.
So, 142857 * 2 = 285714, * 3 = 428571 and so on.
So, to multiply this number by two digit number, let's say 37 - you can do it the following way (math explained):
142857 * 37 = 142857 * (7 * 5 + 2) = (142857 * 7) * 5 + 142857 * 2 = 999999 * 5 + 285714 = 5000000 - 5 + 285714 = 5285709.
So basically, first you divide 37 by 7, and get 5. This is the first digit. After that you put the digits of multiplication by 2 (37 - 5 * 7) and you have 285714. And the last step you subtract the first digit (5) from result.
Related to this topic, I like the trick with number 142857, which you obtain from dividing 1 / 7.
So, besides 142857 * 7 = 999999, you also need to know the following feature: Multiplying by x < 7, the result will be the same number rotated.
So, 142857 * 2 = 285714, * 3 = 428571 and so on.
So, to multiply this number by two digit number, let's say 37 - you can do it the following way (math explained):
142857 * 37 = 142857 * (7 * 5 + 2) = (142857 * 7) * 5 + 142857 * 2 = 999999 * 5 + 285714 = 5000000 - 5 + 285714 = 5285709.
So basically, first you divide 37 by 7, and get 5. This is the first digit. After that you put the digits of multiplication by 2 (37 - 5 * 7) and you have 285714. And the last step you subtract the first digit (5) from result.
Squares of numbers composed of 1's: 11^2 = 121, 111111^2 = 12345654321, 1111^2 = 1234321, etc.
Squares of numbers composed of 3's: 33^2 = 1089, 3333^3 = 11108889, 33333333^2 = 1111111088888889.
Squares of numbers composed of 6's: 66^3 = 4356, 66666^2 = 4444355556, 66666666^2 = 4444444355555556.
Squares of numbers composed of 9's: 99^2 = 9801, 999999^2 = 999998000001 and you get the point.
