Saturday, December 29, 2012

How to Divide Any Number by 91

Since this is the last post of 2012, I thought I'd finish the year with a mental math technique. Before I do that though, I want to show you a pattern that is closely related.

Take any three digit number. For this, I will use 123.


Multiply that by 13. You should get 1599.

Now, multiply that by 11. You should get 17589. You might have done that in your head using the multiplication by eleven trick I taught last October.

Now, multiply that by 7. This will give you 123123. And if you'll notice, we started with the number 123 and finished with two 123s.

Why did this happen? It is very simple, and just relies on the fact that 13 x 11 x 7 is 1001. So, by multiplying by these seemingly random numbers is really multiplying by 1001. And any number times 1001 is just itself repeated twice.

Division by 1001 results in a similar answer. For instance, 123 ÷ 1001 = 0.122877122877...

If you notice, the first three digits are 122, which is just 123 - 1. The next three are 877, and 122 + 877 = 999.

This pattern continues as well. I am not sure how to prove that, but please comment if you do know.

So, to divide by 1001, you just subtract one for the first three digits, and subtract the first three digits from 999 for the next three digits.

However, telling someone you can divide any number by 1001 doesn't sound that impressive. Since 1001 is right next to 1000, people will get very suspicious.

That is why I showed you the multiplication pattern. Dividing by 1001 is basically dividing by 13, then 7, then 11.

123 ÷ 1001 = 123 ÷ (13 x 7 x 11)

So, instead of going all the way to 1001, let's just get part way there by dividing by the 13 x 7, which is 91. We will use 33 as the number we are dividing by, or the dividend.


Our goal is to make it a number divided by 1001, since we know how to do that. That means that the first step is multiplying the 91 by eleven. But, if we multiply the denominator by eleven, we must also multiply the numerator by eleven.

33 x 11
91 x 11

We know 91 x 11 is 1001, so we have the problem we are looking for. What is 33 x 11 though? If you go to the multiplication by eleven post, you will see that it is just 363 (add the 3 + 3, and stick it in the middle). This gives us the problem 363 ÷ 1001.

363 - 1 is 362 and 999 - 362 is 637, so the answer is 0.362637362637...

So, to divide by 91, you just multiply the number by 11, subtract one, and subtract that from 999 to get the answer.

Though it is a lot tougher, you can also divide by 77 (11 x 7) and 143 (13 x 11) with the same principle. You are just multiplying the top number by 13 or 7 instead of the easy 11.

The 1001 pattern is really cool, and something you can turn into a really cool trick. The multiplication by eleven is also a really cool effect. But putting them together into the division by 91 is a really impressive feat of mental math that will fool all of your friends.

Saturday, December 22, 2012

Polynomial Multiplication: Don't Spoil with FOIL

I have been trying lately to keep my posts with less algebraic thinking than they have been, and more simple ideas and patterns. There is still algebra mentioned, but I don't want it to be the bulk of my blog. It really isn't what all of mathematics is based on, which is something I discussed at the TEDx Conference in India a few weeks ago. However, this algebraic concept is a really awesome one. This is actually something that I learned in school, which isn't common on my blog.

One of the key concepts in algebra, especially in quadratic and polynomial units, is how to multiply together two polynomials. For instance:

(x^2 - 4x + 3)(2x^2 - 5)

The main method you learn is FOIL, which is meant for a binomial times a binomial. It is an acronym for First, Outer, Inner, Last; meaning that you multiply the first term in each parentheses together, add that to the product of the outer terms, add that to the product of the inner terms, and then add that to the product of the last two terms in each parentheses.

Here, we follow the same idea, except we just multiply every term by every other term in the other set of parentheses. The product would look like:

(x^2 - 4x + 3)(2x^2 - 5)
2x^4 - 5x^2 - 8x^3 + 20x + 6x^2 - 15
2x^4 - 8x^3 + x^2 + 20x - 15

This might seem a little unnatural to you, since multiplication was always taught with the traditional process of setting up the two numbers vertically and multiplying all of the digits. Though I criticized that method and suggested the Criss-Cross Method a few posts ago, we will follow the rules of the traditional one.

Let's set up these two numbers vertically. I will pretend there is a 0x in the 2x^2 - 5 to make it line up.

        x^2 - 4x + 3
x    2x^2 + 0x - 5

First, we would do the -5 times everything above it, from right to left. Just bear with me and pretend this is a traditional multiplication problem. The only difference is that you don't need to carry, ever.

        x^2 - 4x + 3
x    2x^2 + 0x - 5
 -5x^2 + 20x - 15

Now, we could do it with the 0x, but everything would be zero. So, we will go to the 2x^2 and start. But remember with traditional multiplication that we need zeros to be a place holder. So here, we will have some terms with a zero coefficient to be place holders.

How many of them? Well, it is x to the second, so we need two of them. They will be 0x and 0.

        x^2 - 4x + 3
x    2x^2 + 0x - 5
 -5x^2 + 20x - 15
                 0x + 0

Let's multiply the rest.

                              x^2 - 4x + 3
x                          2x^2 + 0x - 5
                       -5x^2 + 20x - 15
2x^4 - 8x^3 + 6x^2  + 0x  +   0
2x^4 - 8x^3 +   x^2 + 20x - 15

And you got the same answer as before! Pretty cool, right!

Let's try it with this one:

(x^3 + 5x^2 - 8x + 10)(-x^3 + 2x^2 + 12x - 7)

           x^3 + 5x^2  -   8x + 10
x        -x^3 + 2x^2 + 12x  -   7

Okay, this looks really tough. I want you to try it on your own with a piece of scrap paper, and I will write the work down below for you to compare.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7
                                     -7x^3  - 35x^2 +   56x - 70
                      12x^4 + 60x^3  - 96x^2 + 120x +  0
          2x^5 + 10x^4 -  16x^3 + 20x^2 +     0x +  0
-x^6 - 5x^5 +   8x^4 - 10x^3 +    0x^2 +     0x +  0
-x^6 - 3x^5 + 30x^4 + 27x^3 - 111x^2 + 176x - 70

If I did my math right, that should be the answer. If I did it wrong, let me know so I can fix it.

If you did it on your own, you probably noticed that it was a lot of work. Can that workload get cut down? This was a question that was not brought up in school.

The answer to that is yes, with a little help from our old friend the Criss-Cross Method. Do you remember how it works? Let's go through it.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7

First, we multiply those last two terms on the end. 10 • -7 = -70.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7
                                                                         - 70

Now, we do our first little cross. We do (-8x) • (-7) = 56x, and 12x • 10 = 120x. Now, we add those together to get 176x.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7
                                                                176x - 70

Now, we do our first three-way cross. 5x^2 • (-7) = -35x^2, (-8x) • 12x = -96x, and 2x^2 • 10 = 20x^2. If you kept a running total, you probably computed in your head the answer to be -111x^2. Notice that all of the variables have the same exponent on them each time. I find that aspect to be the cool part.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7
                                                111x^2 + 176x - 70

Next, we do the four-way cross. Here is where it gets the hardest. x^3 • (-7) = -7x^3, 5x^2 • 12x = 60x^3, (-8x) • 2x^2 = -16x, and 10 • (-x^3) = -10x^3. If you kept a running total, the math there shouldn't have been too challenging. You should have gotten 27x^3.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7
                                  27x^3 + 111x^2 + 176x - 70

Now, we do the second three-way cross. x^3 • 12x = 12x^4, 5x^2 • 2x^2 = 10x^4, and (-8x) • (-x^3) = 8x^4. Add those up and you get 30x^4.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7
                   30x^4 + 27x^3 + 111x^2 + 176x - 70

Now, we do our last cross, with just two computations luckily. x^3 • 2x^2 = 2x^5 and 5x^2 • (-x^3) = -5x^5. Add those and you get -3x^5.

                                           x^3 + 5x^2  -   8x + 10
x                                       -x^3 + 2x^2 + 12x  -   7
      -3x^5 + 30x^4 + 27x^3 + 111x^2 + 176x - 70

Finally, we do our last computation on the far left. x^3 • (-x^3) = -x^6.

                                              x^3 + 5x^2  -   8x + 10
x                                           -x^3 + 2x^2 + 12x  -   7
-x^6 - 3x^5 + 30x^4 + 27x^3 + 111x^2 + 176x - 70

And if you'll notice, this answer is the exact same one as before. I found the fact that you can multiply vertically fascinating, but the fact that you can apply the Criss-Cross Method as well even cooler.

Saturday, December 15, 2012

Triangular Day: How to Square ANY Number

Today is another triangular day! It is the fifteenth of December and 15 is the fifth triangular number.

Two weeks ago, I talked about how the triangular numbers are one of the many figurative families, which are groups of numbers whose elements form the corresponding equilateral polygon. The triangular numbers are the first of these families, because its elements form an equilateral triangle.

The next of these families were the square numbers, and they are the same square numbers you are thinking of. You can easily find a square number by doing n^2, where you square n to get the nth square number.

Today, I want to continue discussing square numbers, but in more of a fun way. But before we get to the fun part, I want to do a little algebra.

We may remember the binomial theorem from algebra. It said that:

(a + b)(a + b) = a^2 + 2ab + b^2

You also might remember the reverse of this.

(a - b)(a - b) = a^2 - 2ab + b^2

But what if we do one plus and one minus?

(a + b)(a - b) =

It isn't very obvious with the simplification of the other two examples. So, let's just do it out.

(a + b)(a - b)
a^2 - ab + ab - b^2
a^2 - b^2

(a + b)(a - b) = a^2 - b^2

There is the simplification. It is commonly called the difference of two squares, and is an easy way to factor quadratic equations. For example, if you had:

f(x) = 4x^2 - 25

You could factor it easily into:

f(x) = (2x + 5)(2x - 5)

This topic comes up in probably every Algebra II class, but nobody ever notices just a little alteration you can make to it.

Let me move to the subject I want to talk about today, which is squaring numbers. You may know that I perform a mental math stage show called Mathemagics, where I do many feats of mental math. One of the most popular of them is the one where I ask an audience member for a number and I square it in my head.

People always jump to the conclusion that it is either super easy (like a hidden calculator gimmick or an unexplainable gift) or super hard (like a complicated formula that no one would ever be able to learn without extensive training). The super easy explanations are completely inaccurate, and the super hard explanations are also fairly incorrect. They are right that I practiced a formula, but it really isn't that complicated. In fact, you are indirectly taught it in your Algebra II class.

Let's try an example, like 18^2. 18 isn't too bad to multiply by, but what number close to 18 is easier?

20. So, we will go up two to twenty and that means we have to go down two to sixteen.


So, the first thing we do is 20 x 16. That may sound tough, but remember that it is only 2 x 16 with a zero tacked onto the end.

 |   \
18   320
 |   /

We are almost done. All we have to do is add to that the square of whatever number we went up and down. 2^2 is four, so we do 320 + 4 to get 324. And there is the answer.

Let's try a tougher one, say 67^2. How about you try it in your head and see if you figure it out.

Okay, this is hard for a second example. 67 is close to 70, so we go up three to 70 and down three to 64.


Now, 70 x 64 doesn't seem any harder than 7 x 64, but you can pull it off. Rather than trying what you learned in school with right to left multiplication, you have to switch to left to right multiplication. This sounds weird, but it is the correct way to approach it mentally.

7 x 60 is 420. 7 x 4 is 28. 420 + 28, left to right, is 448. Tack on a zero to get 4480.

 |   \
67   4480
 |   /

Now what do we do? We add the square of what we went up and down, namely the square of three. 3^2 = 9, so we get:

 |   \
67   4480
 |   /  +   9
64   4489

And there's the answer. For three digit numbers, you do the same general thing. Try 381^2. 381 is 19 away from 400, so we go up to 400 and down to 362.


400 x 362, left to right is 144800, and now we need to figure out 19^2.

  |    \
381   144800
  |    /

19 is one away from 20, so we go up to 20 and down to 18. 20 x 18 is 360 plus one squared is 361.

  |    \
381   144800
  |    /    + 361
362   145161

And there is the answer. With just a couple weeks of practice, this will become second nature.

But why does this work? If you'll remember from our brief Algebra II review, it was right in our face.

a^2 - b^2 = (a + b)(a - b)

We just have one tiny alteration to make. Let's add b^2 to both sides.

a^2 = (a + b)(a - b) + b^2

And there you go. The number we are squaring is a, and the number that we go up and down is b.

Lots of the stuff I put here is stuff that I say you can easily implement into school curricula, but this is probably the easiest thing to put in. All the teacher has to do is mention that if you add b^2 to both sides, you have yourself an easy squaring formula and maybe demonstrate it.

This was very abnormal compared to the other triangular days. But if you read the post of any triangular day, I would suggest reading this one.

If you'd like to see me actually squaring numbers, you can find video footage on my website, or my greatest hits video which I have posted on the blog.

Saturday, December 8, 2012

Game Theory Made Easy: The Criss-Cross Method

A couple weeks ago, we discussed the Criss-Cross Method for multiplication as a way to make it much easier when we get into larger numbers. The traditional method seemed easy, but in reality, it wasn't so easy.

Before I begin, let me mention a quick thing about game theory. Back when I did my four posts on finding game theory strategies, I gave examples of what are called non-zero sum games. These are games where the two players' payoffs do not add to a consistent sum. For instance, in the game between the police and the criminals, we ran across these payoffs.

CrimeLay Low
3, -5
0, 1
-2, 3
2, 0

If you look in each box containing numbers, you will see that the sums are all different.

3 + (-5) = -2
0 + 1 = 1
-2 + 3 = 1
2 + 0 = 2

However, any game you can quickly think of probably does have a consistent sum. Though the card game war is not a true mathematical game (it is not strategic), I will use it for an example.

Every time you put down a card, you either have a card that is higher, lower, or equal to your opponent's card. If it is higher, you keep your card and win their card (a payoff of 1) and your opponent loses their card (a payoff of -1). Same goes vice versa, you lose your card (a payoff of -1) and they win your card (a payoff of 1). If it is a draw, you do war, which makes you put down four more cards in addition to the original one; three face-down cards and one that you use for the actual war. If you win the war, you win five cards (a payoff of 5) and your opponent loses five cards (a payoff of -5), and vice versa.

If you'll notice, in every instance, your payoff is the additive inverse, or the negative of your opponent's payoff. This seems to be rare, but most games are actually like this. These are called zero-sum games, because in every scenario, you and your opponent's payoffs sum to zero.

Back to what I wanted to say today, game theory has something as well that is completely impractical in certain situations. Namely, the method to find mixed-strategy equilibria. It is already fascinating, but the game theory Criss-Cross Method is much more efficient.

Let's take the following fictional example. Say Bob and Joe are playing a game of tennis, and Bob is serving. Bob can choose to serve to the left side of the court or the right side of the court. Joe can choose to position his body weight so that he is ready for a serve coming to the left side, the right side, or just wait and see where it ends up.

Bob has a very strong serve going to the left, and Joe knows this. So, these made-up payoffs are what the odds (in percent form) are that Bob and Joe will win the match:

Prepare LeftPrepare RightWait Until Serve
Serve Left
60, 40
90, 10
       80, 20
Serve Right
80, 20
30, 70
       50, 50

These don't look like zero-sum payoffs. However, I mentioned earlier that a zero-sum game is a game whose sums are consistent, but they actually don't have to be zero. In this case, the payoffs sum to 100.

The next thing we would have to do now is narrow this down from a 2xn game to a 2x2 game. There is a graphing method that I don't have the time to explain right now that will narrow it down for us. The basic purpose of it is to see if there are any of Joe's strategies where no matter what strategy Bob uses, there will be another strategy better than it. In this case, Joe is always better off preparing one way than just waiting until the serve. So, "Wait Until Serve" is a dominated strategy, or it is a strategy that can always be beaten by another one.

Just to simplify the game a little bit, I will divide each number by ten and then subtract five from them. This will make it more like a common zero-sum game.

Prepare LeftPrepare Right
Serve Left
1, -1
4, -4
Serve Right
3, -3
-2, 2

A zero-sum game in this notation can be written with just player one's, or Bob's, payoff in the boxes, and then the analyst can conclude the other player's payoff. Let's write it that way for simplicity's sake.

Prepare LeftPrepare Right
Serve Left
Serve Right

Now, we would normally go about our regular algebra. Let me do it out here:

Bob's Strategy:

(-1)x + (-3)(1 - x) = (-4)x + (2)(1 - x)
-x - 3 + 3x = -4x + 2 - 2x
2x - 3 = 2 - 6x
8x - 3 = 2
8x = 5
x = 5/8

Serve to the left 5/8 of the time and to the right 3/8 of the time.

Joe's Strategy:

(1)x + (4)(1 - x) = (3)x + (-2)(1 - x)
x + 4 - 4x = 3x - 2 + 2x
4 - 3x = 5x - 2
4 = 8x - 2
6 = 8x
6/8 = x

Prepare left 6/8, or 3/4 of the time and right 2/8, or 1/4 of the time. Never make your decision after the ball is served.

Okay, that was complicated. Imagine if we never simplified the game. But the Criss-Cross Method would have made this a lot easier. For Bob's strategy, we would find the absolute value (distance from zero) of the difference of his possible payoffs for serving left. We would do the same thing for his payoffs for serving right.

Serving Left: | 1 - 4 | = | -3 | = 3
Serving Right: | 3 - (-2) | = | 5 | = 5

If we add those together, we get the common denominator used for their probabilities. 3 + 5 = 8, so the probability will be out of eight.

How do we find the numerators? Simply flop the three and the five, or criss-cross them. This will give you 5/8 for serving left and 3/8 for serving right, which is the same as the algebra.

Same thing for Joe, except we go vertically now.

Preparing Left: | 1 - 3 | = | -2 | = 2
Preparing Right: | 4 - (-2) | = | 6 | = 6

Add these together, and we get a common denominator of eight. Now, criss-cross the 2 and 6 to get 6/8 of the time preparing left and 2/8 preparing right. Same as before.

It might seem a little weird at first as to why on earth that would work, which I don't have a clue about. Please comment if you do know why it works.

I probably made it look more complicated than it is with the example I used. However, I wanted to keep the post practical, and you won't find people talking about payoffs of four in life. In fact, you probably won't hear the success statistics in real life. A game theorist would have to take data of both players and determine the statistics. However, the process after must be completed before you actually get down to the criss-cross method at the end. But it is a really cool way to finish off the problem.

Saturday, December 1, 2012

Triangular Day: A Second Figurative Family

Happy December, and happy triangular day! It is the first of December, and one is the first triangular number. One is also a square number, which is a good reason to bring up this topic today.

The triangular numbers are a type of "figurative family," which are sequences of numbers whose elements are numbers that if you draw that many dots, they can form an equilateral polygon (a shape with no curves or openings whose sides and angles are all equal in measure), which is the most common type of figure. For instance, the triangular numbers are a sequence of numbers whose elements form a equilateral or equiangular triangle. This is the first of the figurative families (you cannot draw a one or two sided polygon).

Today, we will turn to the next figurative family, which would be the family whose elements form a equilateral quadrilateral.

Think for just a moment about that. Do you ever hear kids in math class finding the area or equilateral quadrilaterals, or equilateral quadrangles, or equiangular quadrilaterals or quadrangles? I would be surprised if they do, because there is a very simple name for that shape. I defined an equilateral polygon earlier as a shape with no curves or openings whose sides and angles are all equal in measure, and a quadrilateral is simply a four-sided polygon. So, a equilateral quadrilateral is basically:

A four-sided shape with no curves or openings whose sides and angles are all equal in measure.

Do you recognize this definition from math class? It is the exact definition of a square. So, a equilateral quadrilateral can simply be called a square. Because of that, instead of calling the next figurative family the equilateral quadrangular numbers or something, we can call them the square numbers.

Wait a second! The name "square number" is already taken by the numbers whose square root is a integer. The square numbers are 1, 4, 9, 16, 25, and so on. We can't have two sets of square numbers.

But what are the elements of this new figurative family? You can form a 1x1 square with 1 dot, a 2x2 square with 4 dots, a 3x3 square with 9 dots, and so on. We are forming square numbers!

Believe it or not, the square numbers that we normally think of are the next figurative family. And they all form a square, whose side length is the square root of that number.

Before I go into a pattern with square numbers, I want to mention one quick thing that will be important when we study square numbers further. The explicit formula for triangular numbers was essential again and again. So, I want you to think for a moment about what the explicit formula for square numbers would be.

Don't think too hard! It is simply n^2. That is the definition of square numbers, so we can use it as its formula. We may have to complicate it in a later post with something like 2n(n+0)/2, but for now, it is perfect to leave it as n^2.

Okay, let's look at a pattern now. Take two random numbers, like 2 and 3. Find the sum of the square of each of those numbers.

2^2 + 3^2 =
4 + 9 =

Now, double that.

13 • 2 = 26

Can you find a sum of two square numbers that equals 26? This is a pretty easy one, namely the sum of 25 and 1, or 5^2 and 1^2.

Let's try a harder one, the sum of the squares of 4 and 6.

4^2 + 6^2 =
16 + 36 =

Double 52 and you will get 104, which is the sum of 4 and 10, or 2^2 and 10^2.

Let's try a big one. The sum of the squares of 27 and 41.

27^2 + 41^2 =
729 + 1681 =
2410 =

Double 2410 and you will get 4820, which is the sum of 4624 and 196, or 68^2 and 14^2.

I found it amazing that you can square two numbers, double their sum, and find a sum of two new square numbers. I was almost skeptical about a proof for it, but it actually isn't that bad.

In each of the triangular number proofs (except for the induction one), we found a correlation between the number(s) we started out with and number(s) we finished with. This was essential to turn it into a format that can be worked with.

In these three examples, the original numbers that were being squared did have a correlation with the square roots at the end. Those square roots were actually the sum and the difference of the two starting numbers. So, we can now label everything.

Let a = the bigger original number
Let b = the smaller original number

We can then conclude that the square roots are a+b and a-b.

Let's write this as an equation. We are doubling the sum of the squares of a and b to get the sum of the squares of a+b and a-b.

2(a^2 + b^2) = (a + b)^2 + (a - b)^2

Similar to the triangular number proofs, we will multiply each side out and see what we end up with.

2(a^2 + b^2) = (a + b)^2 + (a - b)^2
2a^2 + 2b^2 = (a + b)(a + b) + (a - b)(a - b)
2a^2 + 2b^2 = a^2 + 2ab + b^2 + a^2 - 2ab + b^2
2a^2 + 2b^2 = a^2 + a^2 + 2ab - 2ab + b^2 + b^2
2a^2 + 2b^2 = 2a^2 + 0ab + 2b^2
2a^2 + 2b^2 = 2a^2 + 2b^2

And we ended up with the two sides being equal! What I find kind of cool is that you can now easily create a brain teaser with this, asking something like taking the number 123 and 456 and asking people what two numbers can be squared and then added together to get double the sum of the squares of those two numbers. With this proof, you can easily determine that the answer is 579 and 333. And if you remember from last December, 333^2 is a pretty easy one to figure out.