Math Awareness Month Part 1: Magic Squares

I explained a bit in my last post that April is Math Awareness Month, as well as linked to the poster on www.mathaware.org. In honor of this occasion, I plan to make my posts this month relevant to the pages on the website and the mathematicians hosting them.

April 1st was a day on magic squares, and I am honored to have been the host of that page. There is a recent performance of me doing it, tutorials on how to make various magic squares, and different activities and questions that can further your magic square experience. Click here to see the page.

The Magic of Algebra

Think of a number from one to ten.
Multiply that number by two.
Add thirteen.
Subtract seven.
Divide by two.
Subtract your original number.
You are thinking of the number three.

This is a classic mind reading trick that many people will learn to do when they are little. The premise is simple; the answer is always three.

Why does this work? As a young child, you probably think back to the arithmetic skills you learned in school and realized that you can't prove this trick because you don't know what number they'll choose.

This is a good time to note that when you learned this trick, the first thing you thought was "why?" Many people see proofs as a tedious process that could be avoided just by accepting things to be true. However, that why spark that went off when you learned the trick is where proofs can be applied. I use proofs so often in my posts primarily for that reason: to foster that why spark.

Anyways, this situation is where you need to define that unknown quantity of the number the spectator is thinking of. How do you define an unknown quantity? Well, this is when you pull out a little tool called algebra. You might not expect to apply algebra to a practical situation, but this is when you can. With algebra, we can stop saying "the number that they are thinking of" and start saying "n."

n

Now, the first step was to multiply the number by 2. n times 2 is pretty simple; it is 2n.

n • 2
2n

Next, we had to add thirteen. That can't be simplified much.

2n + 13

Now, we subtract seven. 13 - 7 is 6, so we can simplify it to 2n + 6.

2n + 13 - 7
2n + 6

Now, we have to divide by two. 2 and 6 both have a factor of two in it, so we will not need any fractions.

(2n + 6)/2
2(n + 3)/2
n + 3

The last step was to subtract the original number. But, we already defined that number to be n, so we just subtract n.

n + 3 - n
3

As you see, the n's cancelled, and you ended up with three. Basically, you undid all of the previous operations to end up with one number.

I think that this is a perfect way to show young children magic, but also introduce them to concepts like algebra and proofs by lighting that spark in such a fun and interesting way.

Invisible Dice Trick

Today happens to be the fifth Saturday of the month. Since this doesn't happen very often, I decided to treat you with a little mathematical magic trick.

I first found this trick on the show ScamSchool, where I also found out about Benford's Law. Rather than explaining the trick, you can go watch the video and then I will explain some of the mathematics behind it. Click here for the episode.

This trick requires a little mental math, and also a little algebra to prove to work. It is extremely basic algebra though, so don't get too worried.

With algebra, you denote unknown quantities with a letter. In this example, the unknown quantities are the values of the two dice. We can denote them with a and b. Assume that a is the one that they pulled back.

The instructions were to first multiply a by 2.

a • 2
2a

Next, the spectator had to add five.

2a + 5

Then, the spectator multiplied this quantity by five. This requires the distributive property.

(2a + 5) • 5
(2a) • 5 + (5) • 5
10a + 25

Finally, the person was asked to add the value of the other die, which we called b.

10a + 25 + b
(10a + b) + 25

After rearranging the numbers, you can see the breakdown. If you know that multiplying a number by ten gives you the same number with zero tacked on, you will realize that 10a+b is just a number with a as the first digit and b as the second digit.

So, we have the number that tells us a and b, and this is added to 25 to get the grand total. That is why the performer must subtract 25 to find this number.

I think that this trick is a super simple way to apply mathematics to magic.

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

511

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

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

2(2^n-1)
2^n-1+1
2^n

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!

It's Math in a Card Trick!!!

I haven’t wrote about many mathematical card tricks in the last little while. The last one I remember was the guessing one where you must try to guess each card’s value incorrectly through the deck. Because of probability and the number e, you will find it difficult to get through 52 cards incorrectly.

The reasoning behind that trick is pretty hard to beat. However, that trick definitely isn’t the best trick out there. The one I want to talk about today is a tad more impressive; something that with a little practice, you will be showing to all of your friends.

Here is how the trick goes: take a seemingly mixed deck of cards and cut off a stack. Then, supposedly by feeling the weight of the cards, determine the number of cards in the cut. You can repeat the effect as many times as you like with your own variations (riffling the cards and saying you can hear each click, etcetera), and it should fool your spectators.

The method is extremely simple. It has a little mathematical reasoning behind it, and involves a small mental calculation.

First, you must arrange the deck in a specific order before the trick. You can fan the deck out and people won’t tell a pattern, but there is one. The stack goes like this:

3C

6H

9S

QD

2C

5H

8S

JD

AC

4H

7S

10D

KC

3H

6S

9D

QC

2H

5S

8D

JC

AH

4S

7D

10C

KH

3S

6D

9C

QH

2S

5D

8C

JH

AS

4D

7C

10H

KS

3D

6C

9H

QS

2D

5C

8H

JS

AD

4C

7H

10S

KD

If you look closely at this pattern, it is going up by three for each card. For the suits, you could technically do it however you want for this trick. However, it is traditionally ordered at clubs, hearts, spades, diamonds in a repeating sequence. You can remember this with the acronym CHaSeD.

Once you’ve got that done, you can begin the trick. If you know some false shuffles (the performer seemingly shuffles the deck without changing the order), you can use them. I would not recommend trying a false cut however, since cutting the deck is okay for this trick.

When you are cutting the deck, occasionally glance at the bottom card (do this while making a hand gesture or some natural movement that will not make it apparent that you made that glance). Wait until it is a higher card, and once it is, memorize that card value. Say it is the Queen of Spades.

Now, you can cut the deck yourself, or have the spectator cut it. Once you have the stack, glance at the bottom card. Let’s say that is the Nine of Spades. Here is what you have to do.

First, make sure you understand the following:

A = 1

J = 11

Q = 12

K = 13

So, the Queen of Spades is equivalent to 12. Subtract the 9 (Nine of Spades) from the 12 (Queen of Spades).

12 - 9 = 3

Now, multiply this number by 4.

3 x 4 = 12

Now, ask yourself this question: is there any way I am holding 12 cards?

You will be able to tell if you are or aren’t holding 12 cards. In this case, you would be holding 12 cards.

Let’s say the bottom card is the Jack of Hearts and you cut to the Four of Spades. Same thing; do the following calculation:

(11 - 4) x 4

7 x 4

28

Now, ask yourself if you are holding 28 cards. If you actually try this, you will find it clear that you are holding more than 28 cards.

Here’s what you do. With our sequence, there is a four every thirteen cards. Because of this, we can add or subtract 13 to the number we have and it won’t make a difference. So, let’s try it.

28 + 13 = 41

In this scenario, 41 would be a reasonable estimate, and the correct answer.

Let’s try one more; say the bottom card is the Jack of Diamonds and you cut to the Queen of Hearts. So, you would do:

(11 - 12) x 4

-1 x 4

-4!!!!

You can’t have -4 cards. However, we can add 13 as we please.

-4 + 13 = 9

9 + 13 = 22

This stack would have 22 cards.

It will take practice to make the glance at the bottom of the cut and the bottom of the deck more slick, and the mental arithmetic quicker, but it is definitely a cool effect. If you know false shuffles, it will become even better. But most importantly, it is a fun way to bring math into your everyday life.

Probability, The Number e, and Magic all in one

Since I have not done any mathematical magic posts recently, I wanted to make sure we got one in.

Here is a fun trick you can do for your friends that is all based on simple math. Ask them what the odds would be that if you give them a card in the deck, they could guess the value of the card. They should say one in thirteen, or about eight percent.

What about guessing the value wrong. That would be 92%.

Challenge them to get through the whole entire deck and get every single card’s value wrong. Pretty easy right, you only have an eight percent chance of getting it right. Watch their stunned reaction when they get halfway through and get one correct.

The Method: As I said, it is all based on mathematics. To calculate the odds, you must multiply together the odds of getting it wrong every single time. So, you would have to multiply the .92 by itself fifty-two times. This gives you:

(12/13)^52 = .0183...

Basically, there is a 1.83% chance that you will get through the deck getting it wrong every time. I thought that was pretty cool.

If you’re in a mathematical mood, it gets even cooler. Say you did it with the full card rather than just the value. This would give you:

(51/52)^52 = .3643...

Still good odds. But check this out. Remember back when we studied the number e? Well, what is the reciprocal of e, or 1/e?

1/e = .3678...

They aren’t exact, but that is an extremely close estimate, right! This is true because of our formula for e^x. Let’s go back to it.

e^x ≈ (1 + x/n)^n

In this case, 51/52 would be written as:

(1 + -1/52)^52

With this rule, this should be about e^-1, which means the same thing as the reciprocal of e.

You could use this same logic to figure it out for just the value. (12/13)^52 is the same as:

(1 + -4/52)^52

This means that the odds are approximately e^-4, which is 1/e^4. I thought that this trick is cool, but the reasoning behind it is even cooler.

Can you correctly add six numbers?

Since lately, the posts have been a little too complicated for some people, we are going to take it down a notch and do some simple addition. But it's going to be cool.

Okay, just follow along, and add up the numbers in your head. No calculators, or paper, just your brain. They won't be hard. Take 1000 and add 20. Now, add 1030. Got it? Add 1000. Now, add another 1030. And add 20. Did you get 5000? Congratulations!

What is so cool about this? The cool thing is that the answer isn't 5000, it's 4100. It's a problem that takes advantage of the fact that people hear the numbers, don't see the numbers, when they do this problem. Look at the pronunciation of the numbers:

one-thousand and twenty
two-thousand and fifty
three-thousand and fifty
four-thousand and eighty

What do you want to say next? Five-thousand, right! This is because you've said one-thousand, two-thousand, three-thousand, four-thousand. You naturally want to say five-thousand, and this urge is too strong to actually do it correctly.

Bonus: Since that post was a little shorter than the others, I will show you one more cool-ish thing.

Take the last two digits of the year you were born in
Add your age (as of December 31, 2011)
You are thinking of either 11 or probably 111

I was shocked when I heard someone saying how that was so cool, so I thought I may as well post it. But think about it, isn't the definition of age the current year minus the year you were born on? If you add the year of your birthday to your age, shouldn't you get the current year, or the year before it? I thought it wasn't that interesting, but you tell me if you thought it was cool.

Fibonacci Day: Fibonacci Magic Trick

I don’t know if you noticed, but today is a Fibonacci Day! It is December third, and three is a Fibonacci number. We’ve looked at some cool patterns in Fibonacci numbers, but it’s time we learn how to do some magic with them! Let’s look at the Fibonacci numbers, but this time in lines.

Line 1: 1

Line 2: 1

Line 3: 2

Line 4: 3

Line 5: 5

Line 6: 8

Line 7: 13

Line 8: 21

Line 9: 34

Line 10: 55

Line 11: 89

Line 12: 144

Line 13: 233

Line 14: 377

Line 15: 610

Line 16: 987

Line 17: 1597

Line 18: 2584

Line 19: 4181

Line 20: 6765

What do you think the sum is of all of the numbers up until line, say thirteen? I can tell you immediately that it is 609. How?

Wait, we know this! Remember when we were adding Fibonacci numbers? The answer was always two ahead minus one! But we can take it one step further.

Let’s make our own Fibonacci sequence this time, starting with any two numbers we want. If you are doing it on pencil and paper, you might want to stick 1 – 10, or you can make an excel or numbers spreadsheet and do it as high or little as you want. In this case, we’ll start with 4 and 7.

Line 1: 4

Line 2: 7

Line 3: 11

Line 4: 18

Line 5: 29

Line 6: 47

Line 7: 76

Line 8: 123

Line 9: 199

Line 10: 322

Line 11: 521

Line 12: 843

Line 13: 1364

Line 14: 2207

Line 15: 3571

Line 16: 5778

Line 17: 9349

Line 18: 15127

Line 19: 24476

Line 20: 39603

What is the grand total up to line eight, you can do any line. The answer is 315. It’s still the same exact principle, except for one little thing.

We move up two lines, and then subtract line two. It is the easiest of all things to do! No matter how gigantic the numbers are, you can still pull it off. What’s even cooler is that there is no specific line you are adding up to, unlike other methods that only go up to line ten.

Bonus Trick: Make one of these sequences yourself, and make sure you have at least ten lines. Now, divide the last line by the one before it. In this case, we would be doing 39603 ÷ 24476. You should have 1.61, right?

This is the same thing as the golden ratio appearing in the Fibonacci sequence. To prove it, we will actually do something a little different than usual. We will add fractions “badly.” If you were a young kid, how would you guess adding fractions works?

I’d say add the numerators, then add the denominators. Like ½ + ¼ should be ²/₆, or ¹/₃. This doesn’t give you the right answer, but it does assure that the answer is in between the two fractions. In this case, we are using line ten and line nine.

Line nine has its own formula: 13x + 21y, assuming that line 1 is x and line 2 is y. Line ten has formula 21x + 34y. So, we have:

^{21x + 34y}/_{13x + 21y}

This is the same as adding fractions badly. This says that this ratio is between ^21x/_13x and ^34y/_21y.

^21x/_13x = ²¹/₁₃ = 1.61538…

^34y/_21y = ³⁴/₂₁ = 1.61904…

Both of these numbers begin with 1.61, meaning any number in between them will begin with 1.61. This proves that line ten over line nine is always 1.61, a great bonus prediction effect to the trick.

First, Geometry fails us. Next, Arithmetic. Now, it's Algebra's turn...

Last week, we took a universal property of mathematics and watched it fail to work. This week, we will do it again, but with clear, simple algebraic proofs.

A little over a month ago, we proved that 64 = 65. Now, let's take that down a notch, and prove that 1 = 2. We will prove it two ways, one will be a little easier to understand, and one will involve complex numbers, which we worked with around four weeks ago.

Easy Proof: Let's say that a = b. Pretty simple. Now, we'll multiply both sides by a. This will be step one.

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

How about we add a^2 to both sides. This is step two.

a^2 = ab
a^2 + a^2 = a^2 + ab
2a^2 = a^2 + ab

For step three, we will subtract 2ab from both sides.

2a^2 = a^2 + ab
2a^2 - 2ab = a^2 + ab - 2ab
2a^2 - 2ab = a^2 - ab

For step four, we will factor out a two from the left hand side.

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

For step five, we will divide both sides of the equation by a^2 - ab to give us 2 = 1.

2(a^2 - ab) = a^2 - ab
(2(a^2 - ab))/(a^2 - ab) = (a^2 - ab)/(a^2 - ab)
2 = 1

Complex Proof (literally!): This one does involve complex numbers, so it might become a little bit challenging. However, it is pretty easy to understand, as long as you realize that √(-1) = i.

Let's remind ourselves that -1/1 = 1/-1. For step one, let's take the square root of both sides.

-1/1 = 1/-1
√(-1/1) = √(1/-1)

We can now simplify that to give us:

√(-1)/√(1) = √(-1)/√(1)

For step three, we can eliminate all of the square roots and replace them with 1s and is.

i/1 = 1/i

For step four, let's divide each side by two.

i/2 = 1/2i

For step five, how about we add 3/2i to both sides. Strange attempt, but we can go ahead and do it.

i/2 + 3/2i = 1/2i + 3/2i

Let's multiply through by i. That will be our step six.

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

For step seven, we can simplify this whole mess and see what we get.

i^2/2 + 3i/2i = i/2i + 3i/2i
-1/2 + 3/2 = 1/2 + 3/2
2/2 = 4/2
1 = 2

Again, we are ending up with the strange solution of 1 = 2.

Why on earth could this be? Maybe arithmetic and geometry can make mistakes, but algebra! Turns out, algebra is fine. These proofs are fallacies. See if you can figure out which step is incorrect, and then read the below.

Easy Proof's Fallacy: Turns out, step five was a fallacy. Where were we there?

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

We divided both sides by a^2 - ab to get 2 = 1. The problem lies in the a^2 - ab division. Let's look at it. We know that a = b, so let's plug a in for the b.

a^2 - ab
a^2 - a(a)
a^2 - a^2
0

We have ended up dividing by zero. Since this is not allowed in mathematics, we cannot do this step. That means that 2 ≠ 1, or at least this does not prove it.

Complex Proof's Fallacy: Again, our simplification was where the fallacy lied. In this problem, it was in step two, when we separated the square roots. Let's look at it:

√(-1/1) = √(1/-1)

√(-1)/√(1) = √(-1)/√(1)

We have jumped one step too far. This property we just used is only true for positive square roots. Remember, a square root is a number when multiplied itself gives you the radicand, or number inside the square root symbol. If the square root is negative, then it does not hold true all the time.

If that was hard to understand, let me lay it out in more simple terms. Say we have:

√((-1)(-1))

One way we could do it is multiply together the -1s and get 1.

√((-1)(-1))

√(1)

1

But if you used this property, you would have:

√(-1)√(-1)

i • i

i^2

-1

This gave us two different answers, so this property just doesn't hold in these circumstances. There are tons and tons of proofs like this, even some that involve calculus. I think that these false statements are really cool to look at and try to figure out what the mistake is. So fortunately, algebra has not went under yet.

Watch the Communitive Laws fail right before our eyes!!!

A couple of months ago, we were working with infinite series, and even proved that the primes are an infinite series. Today, we will work with another infinite series, known as the harmonic series. It goes like this:

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8...

Let's add this up. Because of its nature, we can tell it won't go to infinity. If you can picture it, it kind of moves over a certain amount, then goes back in between, then a little forward again, and so on, zeroing in on a specific number. In fact, it is zoning in on a number called ln2, which is somewhere around .683. I don't really know the proof, but it involves a little bit of calculus.

However, let's try something else. How about we rearrange the numbers. Let's go for every odd denominator, we do two even ones.

1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 + 1/5 - 1/10 - 1/12 + 1/7 - 1/14 - 1/16...

It is still the same series because we are adding every odd denominator once and every even denominator once. Let's tackle this in chunks. Let's just group together some terms every so often.

(1 - 1/2) - 1/4 + (1/3 - 1/6) - 1/8 + (1/5 - 1/10) - 1/12 + (1/7 - 1/14) - 1/16
1/2 - 1/4 + 1/6 - 1/8 + 1/10 - 1/12 + 1/14 - 1/16

1/2(1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8)

Just with that little adjustment, we have turned the same series into 1/2ln2. What we've just seen is that the little rule we learned back in second grade with the turn-around facts, fact families, fact triangles, all of that failing right before our eyes. In fact, this Communative Law that we learned can fail when dealing with infinite series involving negative and positive numbers.

What I find odd is that you can rearrange this series to get whatever number you want. If you tried hard, you could rearrange this to get π, e, or whatever else you want!! I haven't really looked into this, but it seems pretty cool.

Bonus Proof: While we are watching the Communative Law fail, we should ask a question. How do we know it is true? Why should 7 bags of 4 apples be the same as 4 bags of 7 apples? This proof is so obvious, yet I would never had thought of it! In fact, one of the things I've wondered for a while is why the Communative Law is true.

Think of it this way. Take a 4 x 7 rectangle made up of dots. How would we figure out how many dots there were total? Well, we could say, "there are 4 rows made up of 7 dots in each row," or, "there are 7 columns made up of 4 dots in each column." Both ways, we are finding the amount of dots in the rectangle. Which one is right? They both are, which proves why the Communative Law must be true.

Why Does 64 = 65? Or Does It...

Why does 64 = 65? What kind of a question is that? Any pre-schooler probably knows that 64 doesn't equal 65. Algebra clearly shows that they are not equal. However, geometry might throw us off track.

Let's take a chessboard. It's an 8 x 8 grid, with a total area of 64 square units. I'd like you to make the following cuts in the chessboard, as well as along the 5th row from the top (3rd row from the bottom).

Now, arrange these shapes into a rectangle. Let's check out its dimensions. We have a side that is 5 units long, and a side that is 13 units long. To figure out the area of the rectangle, we would do 5 x 13 = 65.

Not good enough? Make the shapes into a triangle. We have 10 units for the base, and 13 units for the height. To find area, we do (bh)/2. So, 10 x 13 = 130 ÷ 2 = 65.

How is this possible? We have taken a grid with area 64 and just by rearranging the shapes, end up with a grid with area 65. No, there was no human error involved, your cuts probably were very accurate, and even a perfectly straight cut would still give you 65 as your area. So, how is this possible?

Even I completely understand that this is completely invalid. However, I still can't wrap my head around why it is wrong. I've been told that the squares along the cut after you assemble the shape are not valid squares, which is the only thing that seems accurate. However, with this, I just find it fascinating to take knowledge you learned in kindergarten, or even pre-school, is being challenged with this very contradictory proof.

I also saw this as a proof of last week's maneuver with Fibonacci numbers. I am not quite sure of why, but it does make some sense, that you are turning eight squared into thirteen times five. It does work again with a 5x5 or 13x13 grid, as long as you make the correct cuts. 

Region Revenge: In August, I gave you guys a problem called region revenge. The goal was to find its explicit formula. The answer 2^n-1 is incorrect, as the sixth cut can only make 31 regions, the seventh makes 57, and so on. Here is the correct answer:

An = (n^4 - 6n^3 + 23n^2 - 18n + 24)/24

You could have solved this with the techniques we used for the other problems, just by creating a five way system.

More Patterns at CTY: All in one triangle!!

This week is my second week at Johns Hopkins CTY program. We did a really interesting class on Pascal's Triangle and it's beautiful properties. I loved them and would like to share some with you.

First off: Pascal's Triangle

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

If you look, each number is generated by adding the number above it to the number to the left of the one above it. For instace, the 20 in the last row is generated by adding the 10 above it and the 10 to the left of the number above it.

One pattern is found by adding up the numbers in each row. Check it out:

1 = 1
1 + 1 = 2
1 + 2 + 1 = 4
1 + 3 +  3 + 1 = 8
1 + 4 +  6 + 4 + 1 = 16

It's pretty obvious! It's the powers of two! 2^0, 2^1, 2^2, and so on. I think that is pretty cool, right?

Also, I had noticed that each row written as a number has a pattern not as obvious, but still cool. Check it out:

1 = 11^0
11 = 11^1
121 = 11^2
1331 = 11^3
14641 = 11^4

This also continues, and you will understand why in a future post.

Bonus: Pascal Magic

Though I wrote it so I could type it easily, the triangle is generally written differently. If you google it, you'll see what I mean. Anyways, have someone draw a rectangle around the top one and make it as small or big as they want. Then, tell them to add up all the numbers in the rectangle and you can tell them the answer immediately.

All you need to do is subtract one from the number directly under the bottom of the rectangle.

Challenge: If you were to begin the triangle with say two instead of one and did the trick, how do you find the sum? What about three? Or 100?