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.

Saturday, November 24, 2012

Criss-Cross Method: Multiplication Made Easy

In school, you are taught pretty early on how to do multiplication on paper. For instance:

x  8

That would be tough to pull off in your head. However, in school, you learn that you can simply do it in two steps.

1. 7 x 8 is 56. Write the 6 and carry the five.

x  8

2. 4 x 8 is 32. Add the five to that and you get 37.

x  8

And there you go. This works great for small problems like 2x1, 3x1, 4x1, 5x1, or even ones like 2x2 or 3x2. However, what about something like a 4x4. Your paper would look like:

x          6137
+  53532000

This method is basically the distributive property, where you are multiplying each number by every other number. For instance, a 3x3 algebraically would look like:

Let x = the first digit of the first factor
Let y = the second digit of the first factor
Let z = the third digit of the first factor
Let a = the first digit of the second factor
Let b = the second digit of the second factor
Let c = the third digit of the second factor
Let P = the product of the two factors

P = (z + 10y + 100x)(c + 10b + 100a)

Using this traditional multiplication method, this product then looks like:

P = c(z + 10y + 100x) + 10b(z + 10y + 100x) + 100a(z + 10y + 100x)

This looks pretty complicated. What this method then does is takes it one step further by distributing the a, b, and c through their parentheses. The one thing that it leaves factored out is the powers of ten that are coefficients of the a, b, and c. You don't multiply through the 10 and 100 that are in front of the b and the a.

P = [(c)(z) + (c)(10y) + (c)(100x)] + 10[(b)(z) + (b)(10y) + (b)(100x)] + 100[(a)(z) + (a)(10y) + a(100x)]

You might be thinking that you don't do that much work in your head. I mean, you've been doing this since third grade! But think through what you are doing. How are you creating the terms to add up?

The first term is created by multiplying the last digit of the second factor, or c, by each digit of the first factor. There is a multiplication method called partial products which would multiply the c by the 10y and 100x, but the traditional method uses the power of zero rule to just write the (c)(10y) to the left of the (c)(z), and the (c)(100x) to the left of that.

The second term is created similarly by multiplying the second digit of the second factor, which we denoted as b, by each digit of the first factor. However, we must remember to put a zero down first, which was something constantly driven into our heads by our math teachers. This is taking into account the ten in front of the brackets. The power of zero rule states that when multiplying by ten, we can just tack a zero onto the end of the other factor. So, we tack a zero onto the end of this number that was found by multiplying b by every digit of the first factor.

The third term is also created by multiplying the first digit of the second factor, which we called a, by each digit of the first factor. Since we have a 100 outside of the brackets, we must tack two zeros onto the end of this number.

Finally, we add up these terms, which is exactly what the equation does above to get the product.

Is this really how we want to get to our product? Can't we combine the terms in a little bit of an easier way?

Let's try it. Let's fully distribute every single term through. We will get:

P = 1cz + 10cy + 100cx + 10bz + 100by + 1000bx + 100az + 1000ay + 10000ax

How about we treat the coefficients as if they are variables. In that case, we have lots of like terms to add.

P = 1(cz) + 10(cy + bz) + 100(cx + by + az) + 1000(bx + ay) + 10000(ax)

That looks fairly simple, or at least compared to the first method. However, if you think it through, you might suspect that the numbers you are multiplying together will not be in natural places in your actual problem.

I want you to try to test this. Write the following on a piece of scrap paper lying around.

x     abc

Now, the first step is just cz. Draw a erasable line from z to c. You can see the relationship between the first two numbers being multiplied. Now, erase it.

Here's where it may appear to get awkward. (cy + bz). Draw an erasable line from c to y, and from b to z. What is a easy relationship here?

In fact, you have just made a cross. These two completely random terms have formed a cross on your paper. You can erase the cross now.

Try the third terms, which there are a lot of them! Draw lines between c and x, b and y, and a and z. Believe it or not, you will get a three-way cross, which might resemble an asterisk slightly.

If you continue this process with the other two, you will find just another cross on the left, and then a line on the left side of your problem.

Because of these crosses, this method is called the "Criss-Cross Method." With this crossing pattern in mind, let's try an example. How about 143 x 822.

x     822

If you'll remember, we first have a line on the right side of the problem, connecting the 3 and the 2. So, we must multiply three and two, which will give us six. We then place that directly under the 3 and the 2.

x     822

Now, we must draw our first cross, which is also on the right side of the problem. We have a line connecting the 4 and the 2, and a line connecting the 3 and the 2. Remember that it resembles an X, it is not two parallel lines connecting the four to the left two and the three to the right two. It is the other way around.

Let's find the products of the connected numbers. 4 x 2 = 8 and 3 x 2 = 6. If you'll remember from our algebra, we must add together these products. 8 + 6 = 14.

We cannot stick a 14 there, just like we can't do in any other multiplication method. We will have to put the four, and carry the one.

x     822    1

Now, we have our three-way cross. We must do three computations: 1 x 2, 4 x 2, and 3 x 8. Don't try to do them all at once, just do one at a time. 1 x 2 = 2, and add to that the one we carried to get three. 4 x 2 = 8, and add that to the three to get eleven. 3 x 8 = 24, and add that to the eleven to get thirty-five. We can then put the five and carry the three.

Notice that after each computation, we only had to remember one additional number. This is the power of keeping a running total after each step. The whole purpose of this is to make your final result easier and less messy, which keeping a running total will do.

x     822    3

Now, we will do the next criss-cross, which is the 1 x 2 and the 4 x 8. 1 x 2 = 2, which we can add to the three we carried to get five. 4 x 8 = 32, which we can add to that five to get thirty-seven. We will put the seven, and carry the three.

x     822    3

Now, we are in home stretch. 1 x 8 = 8, which we will add to the three to get eleven. Since there is nothing to carry over to, we will just put the eleven there.

x     822    3

Believe it or not, we have just found our answer. We can also continue this up to a 4x4, or a 5x5, or even higher if you want. The only change is the number of criss-crosses you make. For a 4x4, you do your first line, then a cross, then a three-way cross on the right side of the problem, then a four-way cross incorporating all of the numbers in the problem, followed by a three-way cross on the left side of the problem, followed by a two-way cross, followed by that same last line.

This may look a little cumbersome, especially with the gigantic crosses you have to do. However, this method is a lot easier than your traditional multiplication once you start getting into larger factors, and will take much less time to master.

Additionally, it is fun to impress your friends with. If you have them give you a 3x3 multiplication problem, you will be able to knock it out in ten seconds or less, and will not have written a single number on your paper. Since I am known for my mental math, the criss-cross method is something I can use to fool people into thinking I handled the whole problem in my head in one shot, rather than writing down each digit after I compute it.

Saturday, November 17, 2012

Game Theory: The Prisoner's Dilemma

A couple months ago, I posted several times about the fascinating aspects of game theory, which is the mathematics of games. However, the strategies that we played are formally referred to as spiteful; we won by forcing the opponent to lose. In this post, I will bring up a counterargument that brings up a case where playing spitefully will cost you the game.

This is probably the most classic game theory example: The Prisoner's Dilemma. To put it simply, two prisoners are getting charged for a crime that they most likely did together, but the police aren't sure. So, they set up a deal where they question each suspect privately, and they can choose to cooperate (admit to committing the crime) or defect (claim they did not commit the crime). The punishments are as follows:

If one prisoner cooperates and the other defects, the cooperator is let free while the defector must spend thirty years in prison.

If both prisoners defect, the police don't want to risk wasting the lives of two innocent men, so give them each an eighteen month sentence.

If both prisoners cooperate, they will be punished for their crime with a ten year sentence.

If we were to draw a matrix to represent the prisoners' payoffs, it would resemble this:

    Cooperate             -10, -10             0, -30       
Defect       -30, 0     -1.5, -1.5 

The dominant strategy is clearly to cooperate, with -10 and 0 as your payoffs verses -30 and -1.5 as your payoffs. Defecting seems to be insane.

Try playing this game with a friend repeatedly, and you will find that you both are cooperating every time, leading to a rough sentence for each of you. However, wouldn't you both be better off by both defecting?

Let's say there is a point where you stop playing, maybe twenty times. On the last one, you can get away with cooperating and your accomplice will have just earned himself a thirty year sentence. You just won the game. But maybe your friend thinks the same strategy, so you decide to betray him in the nineteenth game. And it goes on and on until you are cooperating from turn one.

There are a few strategies that seem to work well, which I will list below:

Strategy One: Defect until your friend betrays you

This strategy is the simplest of these three, and is by far the most spiteful. You start by defecting, and expect your friend to do the same. However, the moment he cooperates with the police, start betraying him. Never forgive him, as he deserves this punishment. It is hurting your payoff, but is hurting his even more.

Strategy Two: Do whatever your friend's most recent move was

This strategy is known as "Tit for Tat," which is supposed to come from this for that. Your first move is a defection, and from then on, do whatever your friend did the last turn. If he cooperates in turn two, you cooperate in turn three. This will never make you win, but you will always be extremely close, and if you were in a situation where everyone played their strategy against ten other people, you would come close to winning.

It is effective because of three main reasons (four are generally presented, but I lumped two together since they are closely related):

Niceness/Non-Competitiveness - You are starting by defecting, and it isn't striving to beat out your friend. You won't betray them at the end either, which leads to a mutual trust. As I said earlier, you can't win, so people will be happy to play against you.

Retaliation - If someone decides to betray you, you won't let it slide. You will punish them by cooperating the next turn, which will lose them points as well. In a big competition, they will not like this.

Forgiveness - After you punish your accomplice, you are back on good terms very quickly. If they are willing to regain your friendship, you are happy to start defecting with them.

Tit for Tat is personally my favorite strategy, but some people feel it is still too mean. If you are playing someone who is also doing Tit for Tat, but breaks his strategy to get a few extra points, you are going to be punishing each other each turn for the punishment they gave you. It is then impossible to forgive each other.

Strategy Three: Always defect unless your friend crosses the line

Personally, I think this strategy is too nice. You will always defect except for these four situations:

1. Your accomplice cooperates on the first turn. Immediately switch to tit for tat.
2. Your accomplice cooperates twice in three turns. Switch to tit for tat.
3. Your accomplice cooperates three times in twenty turns. Switch to tit for tat.
4. Your accomplice cooperates five times in one hundred turns. Switch to tit for tat.

It is easier to forgive your accomplice, and will eventually retaliate. However, I think the retaliation is a little to nice. Nonetheless, they are all great strategies that you won't find in your everyday game.

Saturday, November 10, 2012

Triangular Day: Sum of the Cubes is the Square of the Sum

Just like last week, today is a triangular day! It is the tenth of November and 10 is the fourth triangular number. Here are the triangular numbers just to remind you.

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, ...

Most of the proofs I've done with triangular numbers just involved some simple algebra. We just would perform the necessary operations to the explicit formula for triangular numbers n(n+1)/2 and it would simplify to the answer we wanted.

Today, I will be doing something very different that can't be proved with this method. Rather than looking at triangular numbers, let's look at cube numbers. Here are the first few:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, ...

Let's just add up the cubes. We will start by adding the first one, then the first two, the first three, and so on.

1 = 1
1 + 8 = 9
1 + 8 + 27 = 36
1 + 8 + 27 + 64 = 100
1 + 8 + 27 + 64 + 125 = 225
1 + 8 + 27 + 64 + 125 + 216 = 441

Do you notice any pattern with the sums? If you look closely, you'll see that they are all square numbers. But not any square numbers. Look again.

1 = 1 = (1)^2
1 + 8 = 9 = (3)^2
1 + 8 + 27 = 36 = (6)^2
1 + 8 + 27 + 64 = 100 = (10)^2
1 + 8 + 27 + 64 + 125 = 225 = (15)^2
1 + 8 + 27 + 64 + 125 + 216 = 441 = (21)^2

They are squares of the triangular numbers. When I first saw this, I was stunned that this happened, and thought there was no way this could continue to work forever. But in fact, we can prove it.

You could try doing it algebraically, but we can't represent cube numbers in a form that can be simplified. For this, we have to use something that I've mentioned before called proof by induction.

We already know it works for the first number. We know that:

1^3 = 1^2
1^3 = (1(1+1)/2)^2

Remember that triangular numbers can be found with the formula n(n+1)/2. Since we are squaring it, let's just simplify that.

(n(n + 1)/2)^2
((n(n + 1))^2)/((2)^2)
n^2(n + 1)^2/4

We know it works at some point, that is when n = 1, we can state our induction hypothesis. This is saying that for some number k, we know it will work.

1^3 + 2^3 + ... + (k - 1)^3 + k^3 = k^2(k + 1)^2/4

Now, we want to find out if we continue forward, it will keep working. In other words, we want to know if it holds true for k+1.

1^3 + 2^3 + ... + (k - 1)^3 + k^3 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4

Based on our induction hypothesis, we know what the sum of the first k cubes is. So, we can substitute that into the new equation to get:

1^3 + 2^3 + ... + (k - 1)^3 + k^3 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4
k^2(k + 1)^2/4 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4

If you look on the left side, you will see that both terms have a factor of (k+1)^2. The distributive property says that we can "factor out" or restructure the equation so that the (k+1)^2 is multiplied at a different time. This gives us:

k^2(k + 1)^2/4 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4
(k + 1)^2 • (k^2/4) + (k + 1)^2 • (k + 1) = (k + 1)^2(k + 2)^2/4
((k + 1)^2)(k^2/4 + k + 1) = (k + 1)^2(k + 2)^2/4

Now, rather than having a denominator on just one term, let's create a common denominator of 4.

((k + 1)^2)(k^2/4 + k + 1) = (k + 1)^2(k + 2)^2/4
(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2(k + 2)^2/4

The two sides are almost identical. The only thing that is different is the (k + 2)^2. But we can simplify that.

(k + 2)^2
(k + 2)(k + 2)
(k)(k) + (k)(2) + (2)(k) + (2)(2)
k^2 + 2k + 2k + 4
k^2 + 4k + 4

Let's substitute that in for (k + 2)^2.

(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2(k + 2)^2/4

(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2(k^2 + 4k + 4)/4

And now the two sides are equal. So, when all of the dust settled, it showed that the sum of the first k+1 cubes equalled the square of the k+1st triangular number, or the sum of the cubes is the square of the sum. In other words, the sequence continued to work from k to k+1 meaning that whatever value k is, the next value will hold true as well.

This proof is very complicated, with factoring, substituting and simplifying. But if you look at it a couple of times, it makes perfect sense and makes the fact even cooler.

Saturday, November 3, 2012

Triangular Day... times eight plus one

I don't know if you noticed, but today is a triangular day! It is the third of November and three is a triangular number. Next week is also a triangular day, and I will be mentioning probably my favorite thing about them.

Let me quickly list some triangular numbers.

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120,...

They are basically the sum of consecutive natural numbers. So, 1 = 1, 3 = 1+2, 6 = 1+2+3, and so on.

This week, I wanted to try something. In August, we tried doing a triangular number times nine plus one. That resulted in other triangular numbers, which was fairly cool.

Today, let's try taking that down a notch. We will multiply by eight and add one.

1 x 8 + 1 = 9
3 x 8 + 1 = 25
6 x 8 + 1 = 49
10 x 8 + 1 = 81
15 x 8 + 1 = 121
21 x 8 + 1 = 169
28 x 8 + 1 = 225

Do you see any pattern? This is a little bit harder than the rest, but look closely.

They are all square numbers! It is fascinating to me that triangular numbers generate squares on their own, but it is true.

1 x 8 + 1 = 9 = 3^2
3 x 8 + 1 = 25 = 5^2
6 x 8 + 1 = 49 = 7^2
10 x 8 + 1 = 81 = 9^2
15 x 8 + 1 = 121 = 11^2
21 x 8 + 1 = 169 = 13^2
28 x 8 + 1 = 225 = 15^2

How do we know this goes on forever? Well, let's try to prove it. We know that the triangular numbers follow the formula n(n+1)/2. You can click here to see the post where that was proved.

Also, notice the numbers being squared. They are 3, 5, 7, 9, 11, and so on, which are all odd numbers. They fit into the formula:

2n + 1.

So, we can set up the following equation, considering n is what term we are using in the sequence. So, for the second triangular number, n = 2.

8(n(n + 1)/2) + 1 = (2n + 1)^2

Now, we will multiply through on both sides, so there are no parentheses.

8(n(n + 1)/2) + 1 = (2n + 1)(2n + 1)
8(n^2 + n)/2 + 1 = 4n^2 + 2n + 2n + 1
4n^2 + 4n + 1 = 4n^2 + 4n + 1

And already, we see that the two things are equal.

I found this pattern especially cool because of the switch between sequences, which Fibonacci numbers do all of the time. Probably next spring, I will be talking about the fact that triangulars and squares are much closer related than you'd think. Just look at their names and you should be able to see what I mean.

Saturday, October 27, 2012

The Best Approach to Math Education

All of the things I have posted on my blog are really cool aspects of mathematics. And they do fit into mathematics curriculums perfectly. For instance, while going over prime numbers in fifth grade, you can teach why they go on forever. It isn't too difficult of a concept, it just requires a couple seconds to explain what a factorial is.

But there are lots of changes that can be made in mathematics education that will enhance not only the fun, but the application as well. In my opinion, there are 3 P's to cool math concepts, that a good chunk of my posts fall under. They are proofs, patterns, and practicality.

I have done numerous posts on proofs and patterns, but very little on practicality. Last week's post was a practicality one; it studied a type of problem that you run into in daily life. True, you hopefully won't run into that situation when you are being held captive by the cops, but certain situations in sports and economics will stick you with that type of choice to make, where the dominant strategy isn't always the best.

Practicality is the center of school education. School is preparing you for the outside world, getting you ready for when you have to pay your taxes, finding a good discount at the mall, or even playing a round of poker. And school puts you through the sequence of Algebra, Geometry, Trigonometry, Calculus, and if you continue forward, you might run into Statistics, Linear Algebra, Computer Science, and Discrete Mathematics.

On the topic of mathematical areas, the survey link at the top of the page is for some data for my presentation coming up in New Delhi, India. If you can take a moment to fill that out, it would be great. And it pertains to branches of mathematics.

Back to what I was saying, is this Algebra/Calculus sequence really what we need to succeed in life? Andrew Hacker, a political scientist at Queens College, published an article taking his stand on the topic. The article is a little long, but it is truly worth the time. Click here to read it.

Vi Hart, a "Mathemusician" who makes some really amazing videos for Khan Academy, recently made a video showing algebra and how it is significant, as well as really amazing. This video is also really cool, whether you read this article or not, but the inspiration for it is clear. Click here to see the video.

Coincidentally, I got to meet her at Gathering 4 Gardner this past March, and I found out about this through some other people I met at Gathering 4 Gardner.

I also have a strong opinion on this case of whether Algebra is necessary to teach in school. Though I don't really have a qualified background like Andrew Hacker and Vi Hart, I do speak for a student who just completed the Algebra curriculum and is watching my peers go through it.

Hacker brings up a very good point that Algebra is not a good model for real life situations. When playing football with friends, you are not calculating the distance of the throw by finding how long the parabola representing the arc made by the ball after the quarterback releases it is before it crosses the x-axis while determining how many yards per second the ball is traveling so you can plug that into d = rt and by solving for t, determine how long it will take for the ball to get to the ground, and then plugging the x-intercept, which took approximating the radicals in the quadratic formula when realizing that the equation could not be factored evenly, in for d to find the rate, which you must then switch your speed to in order to be in the right spot at the right time so you can catch the ball without having to dive or stutter. That is completely crazy. What a good player does is approximates where the ball should land, and tries to judge what speed he must take to get to that spot, understanding that he may have to stutter for a second or lunge forward to make the play.

This is absolutely correct. But also, consider the fact that subconsciously, you are following those exact steps, just without the numbers. You must consider the speed the ball is going, when it will be at a reasonable height off the ground, and how far you have to go in order to receive it. Of course, you are not actually taking data points to construct a parabolic figure and using some distance formula I've never heard of to get a number that will be divided by however many milliseconds it took to get from point A to point B to find the precise velocity of the ball. But we do need to have somewhat of an understanding of this information.

Hacker would rather have every student's big focus be quantitative reasoning: mathematics that can be applied to real-life issues. This also makes sense, as you need a grasp on how to keep your numerical aspects of life under control.

Yet, people already seem to be doing fine. We seem to have a well enough grasp of quantitative reasoning to keep America's middle class strong. Yes, the economy has gone under due to people buying things that they couldn't afford. But even if quantitative reasoning was the class of study, there would still be people that didn't pay well enough attention and not using their money wisely. And just like how people would struggle through algebra, they would struggle through quantitative reasoning.

If people already have enough quantitative reasoning skills to survive in the economy, then we don't have to teach it in school. However, we can get a little more advanced in quantitative reasoning, which will require a little algebra.

Again, I want to allude to the fact that I have no qualifications or experience in teaching mathematics. I do want to propose what I think could potentially work well from seeing news, my peers, and the courses I have taken in math and science.

I do believe that a basic understanding of algebra is somewhat necessary. Aside from the fact that quantitative reasoning will begin to require algebra, there is also a more personal reason. In late middle school to early high school, you are not yet sure of what your career will be. Through a study of your most important algebraic concepts incorporating the things Vi Hart showed us in her video, and the things I post on my blog on a weekly basis which neither are taught in the current curriculums, students might want to continue to study mathematics further. If we are adding some of these patterns and proofs into the classroom, then we are adding a chance of inspiring students to pursue a mathematical job, which requires a calculus sequence more than a quantitative reasoning sequence.

After you have completed your math fundamentals from second to sixth or seventh grade, you can then take your traditional Pre-Algebra or Introduction to Algebra course. Following that can be a Fundamentals of Algebra course, which would involve the basic concepts you cover in algebra:

- a moderately rigorous study of linear equations (just a little less detail than the current curriculum)
- a less rigorous study of quadratics (solving for x and some simple transformations and interpretations)
- an overview of the other four parent functions (radicals, cubics, rationals, and absolute value)
- a very brief overview of trigonometric functions (just a basic idea, no involved studying)
- some basic number theory concepts (types of numbers, sequences)
- a review of probability from previous years
- an incorporation of details that will spark interest in the minds of the students

Rather than covering two full years of algebra, you can put the most important parts into one. You can cut out that month of factoring quadratic equations, the week of conversions between slope-intercept and point-slope forms, and the daily twenty minutes spent reviewing number eighteen from the homework because it required too many steps. Though algebra isn't so necessary throughout regular life, you need a basic understanding for most fields of science, engineering, and technology, as well as for pursuit in math itself. By erasing most of algebra from your mathematics curriculum, it becomes impossible to teach high school physics and chemistry classes while kids are still working through their quantitative reasoning course.

After students have a grasp of the idea of algebra, they could then enter a Quantitative Reasoning course, but not the same exact type as Hacker proposed. This type of course might have:

- some finance concepts that will be critical for life outside of school
- a continuation of the probability concepts from the introductory algebra course
- an overview of game theory with a focus of real-life models and situations
- an overview of economics with a similar focus
- an overview of mathematical logic with a similar focus
- an overview of inductive and deductive reasoning with a similar focus

This type of course will prepare you not only for the situations directly involving numbers, but making rational decisions in whatever field you are in, and starting to think critically, which is a popular goal in the people I know and admire.

After this, a student may be a freshman, sophomore, or junior in high school. By then, they will have an idea if they want to go into a STEM (Science, Technology, Engineering, Mathematics) field, or if they would rather pursue other interests. If they would like, they could follow a calculus type of sequence like so:

Integrated Mathematics A (some more detailed review of algebra, and a thorough geometry course)
Precalculus (some more trigonometry, and a thorough precalculus course)
Calculus (a thorough calculus course)

This would prepare them for a job involving lots of higher level mathematical thinking. Students who do not plan to follow a mathematical career can do a more practical mathematical study, such as the following:

Integrated Mathematics B (some review of algebra, an overview of geometry/trigonometry/precalculus)
Probability and Statistics (a thorough statistics course)
Discrete Mathematics (a course that teaches combinatorics and number theory)

This series of courses would be much more relatable to daily life. The STEM students already learned the fundamentals necessary in Quantitative Reasoning after their first algebra course, but the students with other interests can become more advanced in the practical areas of mathematics. By the way, integrated mathematics is a course that combines Algebra and Geometry concepts that I saw as a one year version of Algebra II/Geometry at Phillips Exeter Academy, which is one of the six high schools I am looking at. It seems like a good idea because while we still use the traditional calculus sequence, at least you are getting through it quicker. This integrated mathematics course is important to keep, because the algebraic and geometric models found in areas like statistics and discrete mathematics must be interpretable. However, it would be much less detailed than the course that I named Integrated Mathematics A, since this thorough understanding of Algebra II and Geometry is not necessary for the jobs that these students would want.

Then, they would get their thorough statistics course, which would prepare them for analyzing or collecting data, as well as teach some more advanced probability calculations. If they get through that, they would be ready to take a Discrete Mathematics course, which is also mathematics that is applicable to real-life situations.

I was not surprised when I saw the reaction of the mathematics community, to immediately try to defend Algebra's case. But we did not jump to defend pi's case when Bob Palais and Michael Hartl presented tau. Yes, algebra has so many cool things and does have some practicality. But just like with pi, math education has to change.

Hackler wanted to take algebra out of your required courses. I disagree with this as well. Without a basic algebra course, students cannot understand lots of the proofs and patterns people like myself talk about, they can't find out if they would like to move forward in a STEM profession, and they can't do as much in their Quantitative Reasoning sequence. However, instead of spending three years on Algebra and Geometry, spend two on it and use that third year to teach Quantitative Reasoning and let the students choose where they want to go.

Yes, I have no qualification or background in teaching mathematics to students. This course selection may not be in the correct order, have the correct names, or even be the correct courses. But as a student watching this whole process happen, I can tell that some students would value from something like this STEM-preperatory sequence while others would have much more of a benefit from something like the other sequence. And everybody gets the best of both worlds.

Saturday, October 20, 2012

In Math, Why is Nothing Something?

You may wonder in math why zero is a number? A number is a quantity, a unit to describe something in. So, why should nothing be considered a number? You can't use it to define distance or volume or weight, you can't assign it positive or negative, it even says in its name that it isn't a natural number. Why is it there?

Back when we found out that √2 isn't rational, we used something called proof by contradiction. We assumed that it was natural, and ran into trouble. Let's try that here. We will create our number line with the zero gone.

       -4   -3    -2   -1     1    2     3    4

First of all, try putting a dot where 1/4 would be. If you actually figured out proper guidelines to do it, plot -1/4 with those rules. If you successfully did this and 1/4 is bigger than -1/4, please explain it in the comments section. I would be curious to know.

The true proof is in the relationship in the numbers though. Find the difference between each number in the number line.

     4 - 3 = 1
     3 - 2 = 1
     2 - 1 = 1
 1 - (-1) = 2
-1 - (-2) = 1
-2 - (-3) = 1
-3 - (-4) = 1

Our intervals weren't correct, since there is a two left there. So, we need to figure out what number we forgot in there...

The zero has to be there! We cannot have a proper number line without it.

If you begin to think about other circumstances, you will realize that you need the zero there always. How do you have the origin of a Cartesian plane, or do the Fibonacci sequence backwards?

Using this method, you can also figure out why there isn't positive zero and negative zero. 0 - (-0) = 0, and there can't be an interval of zero in our number line either.

Bonus: In my game theory class this summer, we had a problem where the twelve of us were prisoners, and our teacher told us that he would free us if we could complete the following task:

He had two sets of fourteen cards, each one with a different number on it from one to fourteen. He also numbered each one of us from one to fourteen. He shuffled the cards, and put each one under a different card. So, card one doesn't have to have card one in it, and so on. He then had us discuss our strategy, and then come in one by one and choose an card. If you saw your number on it, you were finished, and if it didn't, you chose a second card. If you didn't get your number after seven cards, he gave all of the prisoners a strike and then reshuffled the cards. Each person went in one by one, and went through that process. We couldn't communicate at all while we were completing the task, we just waited until it was our turn, went in and chose the cards, and then walked out silently without telling anyone what we saw. After all fourteen of us went, we walked back in to hear that each one of us found our number in seven tries or less, and we were free.

There is no strategy that will guarantee you success, but we found a strategy that gets you very close. I want you to try to come up with a strategy that will give you the best odds possible, and comment it. After a month, I will post the best strategy I saw and the one that our class figured out.

Saturday, October 13, 2012

100th Post: The First Secret of Mental Mathematics

I am thrilled to announce that today marks the 100th post on Cool Math Stuff. It looks like this will also be the longest post, since I have so much I want to say. Anyways, it is really great to be able to talk about what I think makes math cool. But I don't want it to just be what I think, I want you guys to give your input as well. So, please feel free to comment, or send me anything that you find cool about math. I will be more than happy to post about it (it saves me from having to think of something cool to write about).

Before I go into my post, I wanted to start finding out what you think is interesting in math. So, I want you to comment with the most interesting post I have written (the title or link will be fine), and I will take the one with the most comments and either repost it or add some input to it. Here are my top ten favorites, which you can feel free to choose from, or find one of your own.

#1. April 14, 2012; Probability, the Number e, and Magic all in one

#2. August 27, 2011; Do the Primes go to Infinity and Beyond?!!

#3. July 9, 2011; Divide Almost Any Odd Number into a Number Consisting of all Nines

#4. March 24, 2012; Pi, Lie, Same Thing... (Part Two)

#5. September 10, 2011; Another Probability Paradox: What’s Your Birthday?

#6. October 15, 2011; Why Does 64 = 65? Or does it...

#7. June 25, 2011; Greatest Common Factor Made Easy: It's Euclid to the Rescue

#8. September 17, 2011; What does .99999... Really Mean?

#9. July 23, 2011; Patterns and Puzzles at CTY

#10. January 14, 2012; Can you correctly add six numbers?

Since this is the 100th post, I wanted to make it something that is really special. And in fact, it is what pulled me into the fascinating aspects of mathematics.

In February of 2010, I was shown a video on TED of math professor and mathemagician Dr. Arthur Benjamin, who was doing incredible feats of mental math in his head. As a curious ten-year-old, I of course wanted to be able to do it too. So, my dad bought me his book, Secrets of Mental Math (there is now a fantastic video course on it, which you can click here to see), and I was slowly pulled into the whole mathematics, magic, and science community.

But those first few pages were what really drew me in. So, I wanted my hundredth post to be about what those first few pages were on: how to multiply by eleven in your head.

First of all, lots of you probably remember from third grade when you learned the following pattern:

1 x 11 = 11
2 x 11 = 22
3 x 11 = 33
4 x 11 = 44
5 x 11 = 55
6 x 11 = 66
7 x 11 = 77
8 x 11 = 88
9 x 11 = 99

That's always a fun one. But there still is a pattern that continues. Let's move it up to two-digit numbers.

10 x 11 = 110
11 x 11 = 121
12 x 11 = 132
13 x 11 = 143
14 x 11 = 154
15 x 11 = 165
16 x 11 = 176
17 x 11 = 187
18 x 11 = 198

This one is a little harder to see. Let me make it a little easier to see.

10 x 11 = 110
11 x 11 = 121
12 x 11 = 132
13 x 11 = 143
14 x 11 = 154
15 x 11 = 165
16 x 11 = 176
17 x 11 = 187
18 x 11 = 198

You might notice that going down the list without the bold gives you 123456789. But the important part is the bold. What numbers do you see?

The numbers we were multiplying by! So, the first digit is the first number and the second digit is the second number. That takes care of a lot of it. But what about the middle digit? Do you see any pattern between the outside numbers and the inside number?

1 + 0 = 1
1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
1 + 4 = 5
1 + 5 = 6
1 + 6 = 7
1 + 7 = 8
1 + 8 = 9

The two outside digits actually sum to the middle one. Let's try a couple. 26 x 11.

2 + 6 = 8

How about 61 x 11?

6 + 1 = 7

Try 72 x 11.

7 + 2 = 9

What about 39 x 11.

3 + 9 = 12

Wait, how can that be? 39 x 11 can't be bigger than 72 x 11. We must have messed up.

I neglected to mention before that there is only room for one digit in the middle spot. Since the full 12 doesn't fit there, the two drops in and the one carries over to the 3, making the answer 429.


What is 67 x 11?


Try 96 x 11. This might be confusing, but stick to our rules.


Let's look at three digit numbers. Do you see any patterns with these ones:

132 x 11 = 1452
254 x 11 = 2794
816 x 11 = 8976
427 x 11 = 4697
354 x 11 = 3894

For the first one, you see the one and two on the outsides, just like the two-digit numbers. But where is the three? What are the four and the five for?

Well, 1 + 3 = 4, and 3 + 2 = 5. This goes for all of the other ones.

How about we try a couple:

1  5 3
 \ / \ /
16  83

153 x 11 = 1683

7  2  4
 \ / \ /
79 64

724 x 11 = 7964

8  4  2
 \ /  \ /
8      2
12  6_
92  62

9    8     7
 \   /  \   /
9           7

That last one was a little confusing, but you will get the hang of it with some practice. And you can even take this up to four-digits, fives, and more. Let me quickly go over why this works. Let's just do 15 x 11 with classic multiplication from grade school.

x  11

So, 1 x 5 is five.

x  11

1 x 1 = 1. Remember to put the zero down as a place holder.

x  11

Now, 1 x 5 is 5. 1 x 1 is 1.

x  11

Add them up:

x  11

That is great. But, what happened in the process? On the left end, you have just the one (from the one in the fifteen). On the right, you have just a five. In the middle, you have the one and the five, which is the same as adding the two digits together and putting them in the middle.

I find the proof of this pretty cool, but I especially find it cool that you can multiply a number by eleven in seconds with a day or two of practice.

Bonus: I don't normally do a bonus on a long post, but this is something I would really like to share. Recently, I was thrilled to receive an email from a reader of the blog, who wanted to know a formula for figuring out a problem like so (this is his exact words in the email):

"Lets just say I have 30 apples and I want to consume them over a 20 day period.  I want to eat 2 apples a day for as many days as possible; and then cut back to one apple so that I am finishing the last apple on the 20th day."

We can see that you would eat two apples in ten days and one apple for the remaining ten. However, I played around with some numbers and did come up with a formula that will generate the answer to this question.

First, let's set some variables.

a = Total Number of Apples (or object of choice)
d = Total Number of Days (or time interval of choice)
n = Smaller Apple Portion (or object of choice)
x = Number of Days to eat bigger apple portion

To find n, all you have to do is divide d into a and ignore the decimal following. For your example, 30/20 = 1.5, so the smaller apple portion is equal to 1 apple.

To find x, you can use the following formula:

x = a - dn

For your example, you would do:

x = (30) - (20)(1)
x = 30 - 20
x = 10

Ten would be the number of days you eat two apples. To find out how many days you will eat a single apple, just subtract x from the total number of days.

So, if you wanted to eat 50 apples in eight days, you would do the same thing. 50/8 = 6.25, so six is the smaller interval.

x = (50) - (8)(6)
x = 50 - 48
x = 2

You would eat seven apples for two days and six for the rest.

I was really happy to see some feedback and ideas for posts, and I would love for all of you to contribute as well. This is the hundredth post, and I think everything was something that I chose to post. I would really like you guys to email me your favorite cool math thing, and I will be more than happy to post it.

Saturday, October 6, 2012

Triangular Day: Tripled

I don't know if you noticed, but today is a triangular day (and a week away from our hundredth post). It is the sixth, and six is a triangular number. We haven't really tried to prove stuff in a little while, so I wanted to take the time and do that today.

First off, let me tell you what we are going to do. We will prove that any triangular number tripled plus the next consecutive triangular number equals a triangular number.

That was a lot of information there. Let's just do the pattern so you can see better.

1) 3(1) + 3 = 6
2) 3(3) + 6 = 15
3) 3(6) + 10 = 28
4) 3(10) + 15 = 45
5) 3(15) + 21 = 66
6) 3(21) + 28 = 91
7) 3(28) + 36 = 120

That is pretty cool. But proving it is even better. And just like lots before this one, we can prove it with algebra.

Did you notice any pattern in the numbers coming out? Look at the positions of the numbers in the sequence rather than the number itself.

1 ––> 3
2 ––> 5
3 ––> 7
4 ––> 9

It consistently works out to the following:

3T(n) + T(n+1) = T(2n + 1)

(Remember that T(n) means the nth triangular number).

Using the triangular number formula n(n+1)/2, how do we write these three terms?

3T(n) = 3n(n+1)/2
T(n+1) = (n+1)(n+2)/2
T(2n+1) = (2n+1)(2n+2)/2

So, we can write it as:

3n(n+1)/2 + (n+1)(n+2)/2 = (2n+1)(2n+2)/2

Since I don't love working with fractions, let's multiply both sides by two and get rid of any denominators.

3n(n+1) + (n+1)(n+2) = (2n+1)(2n+2)

Now, let's simplify all of these factored terms.

3n^2 + 3n + n^2 + 2 = 4n^2 + 3n + 2
4n^2 + 3n + 2 = 4n^2 + 3n + 2

Both sides simplify to the same thing, thereby proving the pattern to be correct.

We can also prove this geometrically. There were tiles in first grade that we would play with that had different shapes, like squares, triangles, trapezoids, hexagons, etc.

We always would try to create a really big triangle. A shortcut we used was instead of making the second row with three triangle pieces (two facing up and one facing down), we would just use a trapezoid. If we were to take a three-dotted triangle and turn it into a trapezoid, we would just do this:

•  •  •  •  •
  •  •  •  •

You can see the triangles better like this:

•  •    •    •  •
  •    •  •    •

This relates perfectly to this pattern. We are adding to a triangle (the second number) to get a bigger triangle (the total). And what exactly are we adding? Three triangles, which makes a trapezoid.

Let's use some logic here. The triangle we are adding onto has a length of n+1, as it is T(n+1). So, the trapezoid should have a smaller base of n+2.

The full triangle at the end should have a side length of 2n+1, meaning that is going to be the bigger base of the trapezoid.

To fully prove it, we think about the actual construction of the trapezoid. Remember how I said that two triangles would point up and one would point down?

For the smaller side, the two triangles that point towards it each add one dot to it. The one that points the other way adds n dots to it. You can see that better if you look at the diagram above.

1 + 1 + n = n + 2

For the bigger side, we have only one triangle pointing towards it with a side of one. The others add n dots to it.

1 + n + n = 2n + 1

And there is the proof. I found this fascinating because of the pattern itself, as well as the fact that you can prove it many ways, which is common with triangular numbers.

Saturday, September 29, 2012

Learn to Add Over Fifteen Numbers in Under Fifteen Seconds

Since I haven't gone over any mental math tricks in an extremely long time, I wanted to get one in this week.

Before this post, I have done posts involving multiplication and squaring tricks. But today, I am going to talk about some addition.

There are ways of doing mental addition reasonably quickly, but some specific series of numbers (like the Fibonacci series, which I taught a way to add last December) can be added much quicker. So, we are going to need some more fastidiousness when we get our numbers to add.

The series we will be adding is the powers of two. Have your volunteer choose a number and list out that many powers of two. Let's say they say nine.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 =

Dont retrogress yet. All you have to do is double the final number in the sequence and then subtract one.

256 x 2 - 1
512 - 1

And there is your answer. If you have a little more prowess, you could try going up to sixteen numbers.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + 4096 + 8192 + 16384 + 32768 =

If you do the math, you will find that the answer is 65535.

This method is actually fairly easy to prove. Remember back when we did infinite series? Let's write that one we are subtracting as an infinite series.

2^0 + 2^1 + 2^2 + ... + 2^n-2 + 2^n-1 = 2^n - (2^-1 + 2^-2 + 2^-3 + ...)

Now, add that series to both sides. If we write it as an infinite series, we get:

2^n = 2^n-1 + 2^n-2 + ... + 2^1 + 2^0 + 2^-1 + 2^-2 + ...

The left side is already simplified at 2^n. How do we simplify the right side? I may have mentioned in a earlier post (I'm not sure which) how to simplify this type of series. Let me explain it again.

1. Find what number you multiply each number in the sequence by to get the next number in the sequence. Call that s.
2. Call the first and biggest number in the sequence x.
3. Use the formula x/(1-s) to get the total.

In this sequence, we are multiplying by 1/2. 1/2(2^n-1) can be written as 2^n-2.

The biggest number is the 2^n-1. So, we plug it into the formula to get:

(2^n-1)/(1 - 1//2)

Using the law of exponents, we can simplify that to get:


And there is the proof. Though the proof is kind of cool, I found it cool that you could add this many numbers up so quickly. Good luck fooling people with this one!