Cool Math Stuff

Fibonacci Day: Combining Two Famous Sequences into One Cool Math Stuff Post

2012-03-03T12:00:00.000-08:00

Finally, we have reached another Fibonacci day. It is March 3, and 3 is the fourth Fibonacci number. Let’s look at the Fibonacci numbers again, as they may have faded from our memories over the past month.

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,...

There is also another series of numbers that we did look at back in August called the prime numbers. If you don’t remember them, they are natural numbers that are only divisible by one and itself. For a quick review, 5 is prime because it is only divisible by 1 and 5. 6 is not prime because it is divisible by 1, 2, 3, and 6.

Here are the prime numbers:

2, 3, 5, 7, 11, 13, 19, 23, 29, 31, 37, 41,...

These are two famous infinite series that are both very significant in science and mathematics. Let’s see what numbers are in both series.

2, 3, 5, 13, 89, 233, 1597, 28657, 514229,...

That’s interesting. There is one very cool thing about these numbers though, that is kind of hard to find at first glance. It isn’t about the numbers themselves, but their position in the Fibonacci sequence. Here are their positions:

3, 4, 5, 7, 11, 13, 17, 23, 29

Do you see a pattern? They are all prime numbers, except for that 4 at the beginning. The four is the only exception as far as I know to this rule. If you notice another exception, then please comment and let us know.

If you look at it vice versa though, you don’t see this beautiful pattern though. If you look at the second Fibonacci number, we have 1, which is not prime. It is actually called a universal number.

Also, the nineteenth Fibonacci number isn’t prime. While 4181 looks very prime, it is actually 113 x 37.

There are a bunch more of these, which I will list below:

31. 1346269 = 557 x 2417

37. 24157817 = 10877 x 2221

41. 165580141 = 2789 x 59369

53. 53316291173 = 953 x 55945741

Though there are many exceptions, I find it pretty cool that almost every prime Fibonacci number is in a prime position.

Bonus: Here is a challenge for you. We know that there is an infinite quantity of Fibonacci numbers. In August, we proved there is an infinite quantity of prime numbers. But are there infinitely many prime Fibonacci numbers?

This is a question that mathematicians still have not figured out. If you were to figure this out, you would have eternal fame in the world of mathematics. I’d say that’s a pretty interesting fun fact. (You don’t have to do the challenge)

What is a number that's bigger than infinity? Or is there one...

2012-02-25T12:00:00.001-08:00

Last December, we took the time to really analyze infinity, and we ran into a little problem. We didn’t bring it up too much, but all of the infinite series were the same size! To be more specific, they were all equal to alef naught, we called it. Even the number of fractions couldn’t break this barrier. But we haven’t exhausted our toolbox yet!

Let’s look at the set of real numbers between zero and one. Say you had a list and said that it contained all of the real numbers on it. I’ll just put a few down here:

1) .6283185...

2) .2718281...

3) .1618034...

4) .3141592...

5) .1428571...

6) .4000000

7) .3333333...

I claim that there is a number that is not on your list. Of course, you can notice many off the top of your head, like .666666, .1, .2, ,3, and so on, but there is a systematic way to generate a new number. Take that first number, .6283185, the number tau over ten. Let’s change the first digit of the first number to something different. I’ll just keep it simple by adding one.

8. = .7...

Now, look at the second number on the list, .2718281. This happens to be the number e over ten, but that’s a complete coincidence. Let’s change the second digit of the second number by adding one.

8. = .78...

Let’s continue this process. Take the golden ratio over ten, .1618034. Add one to the eight and get 9. Keep going until you have gone through the full list.

8. = .7822614...

And we have a brand new number. Could it have been on the list? No. Why not? Well, let’s prove it by contradiction (actually, it is kind of a proof by logic if that actually exists, but whatever). Say it were the thousandth number on the list. Well, the thousandth digits of these two numbers must be different since that is how the process works, so they cannot be equal. Therefore, you cannot pair up the natural numbers (1, 2, 3, 4...) and the real numbers between 0 and 1.

In our other infinity post, or our alef naught post, I could say, we saw that you must be able to pair each number with a natural number in order for the size of the set to be equal to alef naught. You can’t do that with this list. Therefore, it is bigger than alef naught, and is its own number. We denote this number with the letter C, for continuum

What about the numbers between zero and two? That ought to be double of zero and one. Well, prove it. Look at the graph of y = 2x.

<graph>

Now, draw a laser beam from the origin up to the two line. This beam hits a point on the one line and then a point on the two line. Pair those two real numbers up. Now, draw another beam. Pair these two numbers up. Keep doing this, and you will find that no matter how many beams you do, there is always a number on the one line to match up with the two line.

With alef naught, we tried to pair everything up with the natural numbers. With C, we try to pair everything up with a real number between zero and one. Here, we have paired up the real numbers between zero and two with these numbers, so this set is size C.

What about all of the real numbers? That should do the trick! Well, same thing, graph any function that has a domain (x-coordinates) between zero and one and a range (y-coordinates) that goes from positive infinity to negative infinity, then you’ve paired up the real numbers between zero and one with all of the real numbers.

If that is equal to C, then what is more than C? Well, how about the points in the unit square; the square that has points (0,0), (0,1), (1,1), and (1,0). That must be more than C. But we actually can pair it up. Let’s say your points was the following:

x = 0.acegikmo...

y = 0.bdfhjlnpr...

Pretend that those letters are digits. Say they were the following:

x = 0.123456789

y = 0.987654321

We can pair this with a real number between zero and one, namely:

0.abcdefghijklmnop...

So, the example we had would pair up with this number:

0.192837465564738291

Basically, these two quantities are equal! We could go on forever, or to infinity I should say, with different quantities and test them to be equal to alef naught or C. There are some quantities that actually are bigger than C, but I won’t go into that now.

Infinity is such a cool concept to wrap your mind around, since you can’t define it or explain it, but you can actually start to grasp it after looking at some of these things. I think it even gives you a better handle on numbers, and understanding how many there really are. Considering that you learn about infinity (without actually saying you are) in first grade, it must be pretty cool!

Graphing Calculator Part 3: Games!!!

2012-02-18T12:00:00.000-08:00

For a little while now, I've been mentioning how fun graphing calculators are. However, it's still all Algebra, and lots of people don't look at the fun side of things as much. Well, we have a solution, and it's all in one button.

The button is the "apps" button. Yes, graphing calculators have an app store, so to speak. The only difference is that everything is free! They are all different tools that will assist you in math class.

Click the "probability stimulator" button after you've clicked apps. This will bring up a number of choices; marbles, spinner, coins, cards, numbers, and dice. After you choose one, just hit enter and it will spin the spinner, roll the die, draw a card, flip the coin, etc.

You can use this for one of two things. You can use it to randomly generate data. While practicing the close-together method, or squaring two-digit numbers ending in five, it is always useful to be able to randomly generate numbers. It can also be used to practice graphing scatter plots and bar graphs.

The other thing you can use it for is to play games. It is a lot more fun than it seems to just gather a bunch of friends up and call out a bunch of numbers, and then have the calculator generate some numbers to see whose number comes up first. Though it isn't extremely mathematical, it is definitely a feature of the graphing calculator that I had to share.

Graphing Calculator Part 2: Data

2012-02-11T12:00:00.000-08:00

Last week, we were using graphing calculators to do some graphing of equations, which was pretty simple. However, these things have way more capabilities than that!

We are going to try to use a graphing calculator to take a set of data, and not only graph it but show us an equation to approximate values of the equation.

The first thing to do is set diagnostics on. To do that, hit the catalog button (to do this, hit 2nd and 0), and scroll down with the arrows on the top right to "DiagnosticOn." To speed up this process, you can jump to the Ds by pressing whatever button has a small blue D to the upper right of it. The x^-1 button, also known as the reciprocal button, has the D next to it, so that will make things quicker. Click diagnostics on, then hit enter. It will say "Done" underneath after you have turned it on.

Now, we can put our data into the calculator. To do this, click the "Stat" button, then click edit. This will bring up a chart with L1 and L2 as the first columns. Fill in L1 with your x-values and L2 with your y-values. To keep things simple, we will use the same data:

L1: 1, 2, 3, 4, 5
L2: 1, 3, 6, 10, 15

Now, we will go fix up our window. Last week, we had the following:

Xmin = -10
Xmax = 10
Xscl = 5
Ymin = -10
Ymax = 10
Yscl = 5
Xres = 1

I'd like you to change it to the following, just for this data:

Xmin = 0

Xmax = 7

Xscl = 1

Ymin = 0

Ymax = 30

Yscl = 5

Xres = 1

Now, we will decide what we want our data to look like on the graph. To do this, we first hit statplot (2nd, Y=), then hit enter for plot one. Make sure that your type of graph is the first one, and Xlist is L1 and Ylist is L2. For mark, put whatever you prefer.

If you hit the graph button, you should see the data all graphed. I think that is pretty cool alone. However, we can do better!

This part is a little complicated and has to be done perfectly to work. First, hit stat again, and this time, move right to calc. Now, choose the type of equation it is. To do this, we can use some of the skills we learned from sequences in July and August, by finding common differences. In this sequence, the equation would be a quadratic.

1 3 6 10 15

2 3 4 5

1 1 1

If you don't know this process, don't worry. You can learn it at this post: http://coolmathstuff123.blogspot.com/2011/07/problem-of-week-day-2-week-of-717-723.html. However, it isn't necessary for the graphing calculator. You can either take an educated guess, or if you really don't know, hit PwrReg.

Since this is a quadratic, hit QuadReg, then enter. This should come up with this screen:

y = ax^2 + bx + c

a = .5

b = .5

c = 0

R^2 = 1

This looks like nonsense, but I'll bet we can interpret most of it. We have already mentioned that the base quadratic equation is ax^2 + bx + c. Well, we have that up top, then a =, b =, and c =. These are actually the values that are a, b, and c in the equation. If you want, you can plug them in to get the following:

y = 0.5x^2 + 0.5x

The R^2 button is a little complicated, but I will explain it briefly. It is called the correlation coefficient, and tells you how close the equation is to being a quadratic, or whatever regression you chose previously. In this case, your value is 1 because it is a quadratic. If it were something like .94386, this would be an equation that was close to a quadratic, but not quite there. The closer to 1 it is, the better the data is. If it were .32894, this would be bad data, and you might want to recollect your data.

But it can do better! For the next step, click stat calc again, and hit QuadReg. Now, we need to put in a little code to tell the calculator what we want it to do with the equation. First, hit L1 (2nd, 1). Then, put a comma, and hit L2 (2nd, 2). Now, put another comma, and hit Y1. To do this, you have to hit Vars, move over to Y-vars, then hit function, and Y1. Now, hit enter.

Wait, it just deleted it! It's okay, it gets better. Click Y=. You should see an equation present; the exact same one that it told us! Let's graph it. To do it, just hit graph. We already set ourselves up.

At this point, we should use this equation to approximate, or in this case determine, more data for the equation. Let's say we want to know the 6th value. To figure it out, do what we would have done before. Hit calc, just above trace, then hit value, and type in 6. It tells you that Y = 21.

In school, we learned about graphing calculators, but I would never have expected that they could do something as crazy as this! Considering that some of my friends who don't really like math refer to my graphing calculator as a "game," I think there must be something cool about it.

Since the algebra has been getting a little heavy over the past few weeks, I will bring it down a notch and prove to you that my friends are correct in calling it a game.

Graphing Calculator Part 1: Graphing

2012-02-04T12:00:00.000-08:00

Mathematics is cool in itself with the proofs and patterns, but there are also tools in mathematics that are pretty cool as well. One of which is the graphing calculator. For the majority of February, I will be teaching how to use a TI-83 graphing calculator, and you will be able to use it and appreciate it as much as your iPods and iPhones.

First of all, these things only work on TI-83s and mostly on TI-84s as far as I know. I'm not sure about other types of calculators, so don't get confused with a different type. If you have either of those, take it out now so you can follow along. If you don't, you are missing out on some awesome mathematical technology, but you should be able to get the vive of what a graphing calculator is.

Okay, now to the fun stuff. What does a graphing calculator imply it should be able to do? Graphing, right! And it's as easy as ever with these devices. Let's say you want to graph 3x - 7.

First, we will turn on our calculators (in the bottom left corner). Now, two from the left on the very top is a purple button with the word window on it. Click that.

This screen enables us to choose what we want our graph intervals to be. First, we will choose what we want our X-minimum to be, called Xmin on the calculator. Let's put in -10. The negative button is one to the left of the Enter button, in white. Don't use the blue minus button, that won't work on the calculator.

Now, we have to choose an X-maximum. To keep things consistent, we'll use ten. Use the enter key as if it were a return key.

Next, we must look at the X-scale, labeled as Xscl. This basically makes a little mark at every interval you type in. I am going to put in five, just so there aren't too many of them on the graph. However, if you want to see intervals of two, feel free to put in two. Make sure you don't put a negative there!

For the y-values, we can change them up, but I will do the same as we did for the x-values. -10 for Ymin, 10 for Ymax, and 5 for Yscl. Leave Xres alone as one.

If you click the purple "graph" button now, you will see your Cartesian plane, with little marks at (-5, 0), (0, 5), (5, 0), and (0, -5). These are simply the intervals, which were correctly placed. If you see then at all of the even intervals, then you put 2 in for Xscl and Yscl.

Now, we will graph the line. To do it, click on the purple "Y=" button, which is right to the left of the Window button. It should come up with Y1, Y2, and so on. The curser should be next to Y1, where we will type in the equation. Type in a 3, then X. X will appear if you press the X,T,Θ,n button, which is a black button located one to the right of the green Alpha button. Now, press minus (not negative this time), and then seven. Now, hit graph again.

The calculator should have graphed the line right before your eyes. Pretty cool, huh!

Say you wanted to know what y was when x = 2. There's no need to approximate with your line. Simply do the following:

1. Hit the calc button. To do it, click 2nd, then Trace. This will bring it up for you.
2. Click on the first choice of value. To do that, simply hit enter.
3. The graph should be brought up again with X= in the bottom corner. Type 2 into the calculator (since two is the value we want), then hit enter.
4. The calculator should have come up with Y=-1 in small print to the right of the y-axis. It has also put a little x over the point (2, -1) to show you where it's located on the graph.

The cool thing is that now that you have set up your window, you can graph anything you want. Say you want to graph -0.5x^2 + 5x - 5. Just click Y= and punch it in, and it will come up with the graph. To do the power, you can do one of the following:

1. Click the carrot just above the division sign, and then a 2
2. Hit the x^2 button to the left of the comma

Say you are curious as to what a quartic equation looks like; an equation with degree four, or an x^4 term in it. Try graphing x^4 - 5x^3 + 4x^2 + 6x - 8. It's pretty cool.

You can also look at trigonometric graphs. Try graphing the tangent of x, or tan(x). It's pretty cool!

Here's something I want you to try to graph:

Y1 = sin(x)
Y2 = sin(x+2)
Y3 = sin(x+4)

Those three look really cool together!

Tip: Remember not to use a minus sign for a negative. It will always mess up something that you were working on for a long time.

If you don't own a graphing calculator, but want to check out these graphs, I'd recommend www.graphsketch.com. The picture of the graph is better, but it doesn't have the same freedoms as a graphing calculator, as you will see next week.

Why is the Distributive Law true?

2012-01-28T12:00:00.000-08:00

Remember back in sixth grade or so, when the teacher said, "Okay, here's the deal. a(b + c) = ab + ac." Maybe not. However, this is called the Distributive Law, or the Distributive Property. My class had to do a whole project on it, so it really got jammed into our heads. However, in the midst of this big project, I always was thinking in the back of my head "Why is this true?" And since the proof is so incredibly simple, I had to share it.

First off, let me describe the Distributive Property in further detail. I've used it in the past without thoroughly going over it, so that must get done now. Take this problem:

8(5 + 3) =

We could do it this way:

8(5 + 3)
8(8)
64

Or, we could use the distributive property. This means we distribute the 8 to the 5 and the 3 to get the answer.

8(5 + 3)
8(5) + 8(3)
40 + 24
64

This doesn't seem useful right now, but let's try it with variables and you'll see. Try simplifying the following.

3(x + 4)

We can't combine the x and 4 because they are not "like terms." We can't add apples and oranges, so we can't add x and 4. However, we can use the distributive property.

3(x + 4)
3(x) + 3(4)
3x + 12

And there you go. You may be wondering what I was at that moment; why does it work? I'll bet you can prove it algebraically, geometrically, and every other way, but we're going to prove this logically. Let's take a story problem:

I have eight bags of fruit, each with five apples and three pears inside. How much fruit do I have in all?

One way we could do it is figure out how much fruit is in one bag. To do it, we would add the five apples to the three pears to get eight pieces of fruit. By multiplying by the eight bags, we get sixty-four.

We could also total up the number of apples and the number of pears. Eight bags times five apples is forty apples, and three pears times the eight bags is twenty-four pears. 40 + 24 = 64, which is the same answer as before.

If you think about it, this second method is the exact same thing as the distributive property; we multiplied each term in the parentheses by the number of bags, which is known as the "factor," then added those two totals together. It makes perfect sense, it just takes that little story problem to understand it. This property is so important in Algebra, from FOIL to combining like terms, from radicals to point-slope form. It is endless what you can do with the Distributive Property. Sometime leading up to the problem of the weeks, I will post about something else you can use this property for, which will most likely come in handy in the problems!

For the majority of February, we will be putting together mathematics and technology, and using a calculator to graph equations, scatter plots, and even play, what my friends call "games" on a calculator. Except they throw coins and dice, not live birds that have explosives inside of them...

Fibonacci Day: Proof of Binet's Formula

2012-01-21T12:00:00.000-08:00

I don’t know if you noticed, but today’s a Fibonacci day! It is January 21, and 21 is the eighth Fibonacci number.

Several months back, I presented the explicit formula for Fibonacci numbers; F_n = (a^n – b^n)/(a – b), assuming the following:

a = 1 + √(5)/2

b = 1 – √(5)/2

How on earth does that work? We’re going to use a technique called “proof by induction” to prove it. This is a difficult technique, but if you really try to follow each step, it will all come together in the end.

Fibonacci Identity	F_n+1 = F_n + F_n-1
Original Formula	F_n+1 = (a^n – b^n)/(a – b) + (a^n-1 – b^n-1)/(a – b)
Add like denominators	F_n+1 = (a^n – b^n + a^n-1 – b^n-1)/(a – b)
Communative Property	F_n+1 = (a^n + a^n-1 + b^n – b^n-1)/(a – b)
Distributive Property	F_n+1 = (a^n-1(a + 1) – b^n-1(b + 1))/(a – b)
Substitute for a and b	F_n+1 = (a^n-1((1 + √(5))/2) + 2/2) – b^n-1((1 - √(5))/2 + 2/2))/(a – b)
Add like denominators	F_n+1 = (a^n-1((3 + √(5))/2) – b^n-1((3 – √(5))/2))/(a – b)
Complicate Fractions	F_n+1 = (a^n-1((6 + 2√(5))/4) – b^n-1((6 – 2√(5))/4))/(a – b)
Factor	F_n+1 = (a^n-1(((1 + √(5))/2)^2) – b^n-1(((1 – √(5))/2)^2)/(a – b)
Substitution	F_n+1 = (a^n-1(a^2) – b^n-1(b^2))/(a – b)
Add Exponents	F_n+1 = (a^n+1 – b^n+1)/(a – b)

Q.E.D

And there’s the proof. It is very confusing, but if you read it through a couple of times, it will begin to sink in. When I see proofs like this, it is massive chaos in the middle, but once it all comes together in the end, it just makes it such a cool proof.

Can you correctly add six numbers?

2012-01-14T12:00:00.000-08:00

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.

How did they find the Quadratic Formula?

2012-01-07T12:00:00.000-08:00

A few weeks ago, we were looking at quadratic equations, equations with an x^2 term in it. We learned how to complete the square, which is translating from standard form to vertex form.

Another cool thing I never mentioned that is cool about vertex form is that you can calculate x when the equation is in vertex form. For (x + 3)^2 – 1 = 0, you can say that x = -2 or -4, just by simply manipulating variables.

(x + 3)^2 – 1 = 0

(x + 3)^2 = 1 (add one to both sides)

x + 3 = 1 or -1 (take the ± square root of both sides; use 1 and -1 because 1
and -1 squared are both 1)

x = -2 or -4 (subtract three from both sides, from the 1 and the -1 to get
both answers)

What about for standard form, using the coefficients a, b, and c. Maybe there is a formula we can derive just by following the steps involved in completing the square. Remember our original equation: ax² + bx + c = 0. We’ll do it alongside an example utilizing every step, 2x² – 4x + 6 = 0.

Starting Equation	ax²+ bx + c = 0	2x² – 4x – 6 = 0
Move the constant over to the right hand side, just to make things less messy.	ax² + bx = -c	2x² – 4x = 6
Divide through by a. Normally, we would factor it, but it is the same thing.	x² + ^b/_ax = -^c/_a	x² – 2x = 3
Take half of the new b term and square it.	(^b/_2a)² = ^{b^2}/_4a^2	(-1)² = 1
Add in and subtract out this number. For this, we’ll just add it to both sides, since c is on the other side.	x² + ^b/_ax + ^{b^2}/_4a^2 = -^c/_a + ^{b^2}/_4a^2	x² – 2x + 1 = 3 + 1
Simplify the right hand side. For the first example, we will multiply the first fraction by ^4a/_4a to get a common denominator.	^4a/_4a(-^c/_a) + ^{b^2}/_4a^2 -^4ac/_4a^2 + ^{b^2}/_4a^2 ^{b^2 – 4ac}/_4a^2 x² + ^b/_ax + ^{b^2}/_4a^2 = ^{b^2 – 4ac}/_4a^2	x² – 2x + 1 = 4
Convert the left hand side into squared form.	(x + ^b/_2a)² = ^{b^2 – 4ac}/_4a^2	(x – 1)² = 4
Square root both sides. Don't forget to put our ± sign before the square root!	x + ^b/_2a = ±^{√(b^2 – 4ac)}/_2a	x – 1 = ±2
Isolate the x by adding/subtracting the remaining constant.	x = -^b/_2a± ^{√(b^2 – 4ac)}/_2a x = ^{-b ± √(b^2 – 4ac)}/_2a	x = 1 ± 2 x = -1 or 3

And we have a formula; x = -b ± √(b^2 – 4ac)/2a. This is known as the quadratic formula, the formula that takes ANY quadratic equation and generates its x-intercepts. Not only that, but it doesn't generate false intercepts, and it is very clear when there are no intercepts.

This formula is pretty well-known, but I think it's cool how it is derived. This is a formula that you want to memorize, because it is so important.

Check out my 2011 Greatest Hits Video and learn a pattern as well.

2011-12-31T12:00:00.000-08:00

A couple of posts ago, I presented my greatest hits video for my Mathemagics performance. If you haven’t seen it before, it is what it says, a combination of math and magic. Now that 2011 has just twelve hours left, it might be a good time to present my 2011 greatest hits video.

For some upcoming shows, I will be at Gathering for Gardner early spring in Atlanta, Curious1729 Con in St. Louis, and many other exciting performances, listed on www.EthanMath.com.

As you probably noticed in the Mathemagics performance, squaring numbers is a big part of my show. I’ve already shown how to square 2-digit numbers that end in five, as well as the close-together method (which works for squaring as well), but there is a certain type of 2-digit number, or any digit number for that matter, that follows a specific pattern. Let me show you:

3^2 = 09 (or just 9)
33^2 = 1089
333^2 = 110889
3333^2 = 11108889
33333^2 = 1111088889

6^2 = 36
66^2 = 4356
666^2 = 443556
6666^2 = 44435556
66666^2 = 4444355556

9^2 = 81
99^2 = 9801
999^2 = 998001
9999^2 = 99980001
99999^2 = 9999800001

See the pattern? This only works for numbers that are all 3’s, 6’s, or 9’s, but it is pretty cool! It’s not a very impressive effect to do for someone, but a really cool pattern, and trick.

Algebra + Geometry + Trigonometry + Arithmetic = A Perfect Cool Math Stuff Post...

2011-12-24T12:00:00.000-08:00

A few weeks ago, we saw the second and third cube roots of one, which used those complex numbers, with i in it. While proving that their cubes were one, we learned how to do some operations with them using algebra. However, there is a really cool geometric way to do the math as well.

First off, take the Cartesian Plane. We'll label the x-axis with the integers and the y-axis with the imaginary numbers.

To plot a point, just move the constant to the left/right and the coefficient up/down. So, for 2 + 3i, you would move two to the right and three up. For 4 - 8i, you would go four right and eight down.

Let's say you had to add these two numbers together. First off, you could do it algebraically, which isn't that cool, and you get 6 - 5i. However, you could also add them on the imaginary plane.

Let's plot the points:

How about we just see what happens when we plot 6 - 5i. We'll connect all of the points to make a little quadrilateral.

What do you notice? The points have formed a parallelogram, with the opposite sides equal in length and parallel. At this point, we should go over how to figure out the lengths.

To figure out the lengths, we use the Pythagorean Theorem. Remember back to the problems of the week? We saw that a^2 + b^2 = c^2? We will be proving that in a later post, believe it or not. Anyways, we will be doing exactly that, literally! The form for a complex number is a + bi, so we can just plug those in and solve for the length c.

In the case of 2 + 3i, we will do 2^2 + 3^2 = c^2.

2^2 + 3^2 = c^2

4 + 9 = c^2

13 = c^2

3.606 ≈ c

To keep things more simple, we will use two different points; 8 + 6i and -3 + 4i. In this case, the lengths are 10 and 5.

Of course, we could add them together pretty easily with algebra or geometry to get 5 + 10i. However, let's figure out the angles of the lines. To do it, we use trigonometry, which we also used for the problem of the weeks.

To briefly review/explain, we will take the b term (which is the opposite side) and divide it by the c term, or hypotenuse, to get the sine of the angle. To retrieve the angle, press the sin^-1 button.

In this case, here are our angles:

8 + 6i: 37° (approx.)

-3 + 4i: 127° (approx.)

Now, we will learn how to multiply them together. All you have to do is two easy steps: multiply the lengths and add the angles. For this one, we multiply 5 and 10 and add 37 and 127.

5 x 10 = 50

37 + 127 = 164

If you do some trigonometry, you will get an approximate answer of 48 + 14i, which happens to be the correct answer.

You can also easily multiply complex numbers by real numbers; just multiply the length by that number. If you think about it, this corresponds to the original method. These are just a handful of the many cool things you can do with the imaginary plane. To be honest, I really love algebra and arithmetic, but I'm not one of those people who is all over geometry (I'm not Archimedes, the famous Greek mathematician who died doing geometry in the sand), but this little method is just so cool.

To infinity and beyond!! With numbers I mean...

2011-12-17T12:00:00.000-08:00

We have done a lot with infinite series already, we used it to supposedly prove the Communitive Property wrong, to practice systems of equations in the problem of the week, and even prove that the prime numbers fall into this category. However, we haven't really looked at infinity. Is it a number? Can we define it? What is greater than infinity?

Let's look at an infinite series. Take the natural numbers; 1, 2, 3, 4, 5, 6, 7, ... This sequence takes us out to infinity. How about 10, 20, 30, 40, 50, 60, 70, ... This takes us to infinity also. However, which sequence has more terms in it?

One side of you is saying, for every term in the multiple of ten series, you need ten terms from the natural number series to get that high. Therefore, the first series must be ten times bigger. The other side of you is saying, for both series, we are going out forever. This means they must be equal. When I first saw this, I was leaning greatly towards the first side. However, this isn't quite right.

For each number in the first series, we can pair it with a number in the second. For example, we can pair 1 with 10. Then 2 with 20. Then 3 with 30, 4 with 40, 5 with 50, and so on. If the first side is greater, then you must run out of terms on the second side. However, the second side is infinite also. Is the second side were greater (which I cannot make an argument for that), then we would run out of terms on the first side. This means that the sides must be equal.

In fact, any series is the same size as the set of natural numbers if you can write it out with no infinite gaps in between. For instance, the integers:

... -3, -2, -1, 0, 1, 2, 3 ...

You can rewrite this to get:

0, 1, -1, 2, -2, 3, -3, 4, -4...

Since there are no infinite gaps here, that must mean that it is equal to the set of natural numbers.

What about the fractions (you could say the rational numbers, but rational numbers are simplified and fractions can or cannot be)? It's a big statement, but we can try it.

1/1	1/2	1/3	1/4	1/5
2/1	2/2	2/3	2/4	2/5
3/1	3/2	3/3	3/4	3/5
4/1	4/2	4/3	4/4	4/5
5/1	5/2	5/3	5/4	5/5

This is a table of the fractions. Is there a way to write this table without any infinite gaps? Turns out there is. If we draw lines diagonally, and put it all together, it will have no infinite gaps. It would look like this:

1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3, 1/4, ...

This means the amount of fractions is the same! This quantity is denoted with the hebrew letter alef, also called alef zero, or alef naught. However, do you think there is another type of infinity, or is every set equal? I will come back to this in a couple months, and we will find the answer and why.

A Few Awesome Properties of Quadratics

2011-12-10T12:00:00.000-08:00

In the problem of the week, I would occasionally throw in some quadratics, or problems with squaring involved. If you’ve seen my performances, you might have noticed I like things that have to do with squaring, and quadratics is definitely one of them. Quadratics is normally done in the form f(x) = ax^2 + bx + c, which looks nice, but isn’t so useful. However, there is a form that is pretty commonly used that is almost magical in a way.

This form is the form f(x) = a(x - h)^2 + k. First, I will show you how to get into this form, then we will look at its properties.

The technique we will use is called “completing the square.” The first thing you do is factor the a out of the equation. Let’s use 1/2x^2 + 3x + 5 as an example.

1/2x^2 + 3x + 5

1/2(x^2 + 6x + 10)

Next, plug that new b term into b^2/4. What I find easier is to divide the b term by two and then square it, so the division doesn’t get too messy.

6/2 = 3

3^2 = 9

What this means is that the equation (x^2 + 6x + 9) is square, or a “perfect square trinomial.” It is in fact (x + 3)(x + 3). To figure out that three, all you have to do is divide that b term by two.

There is only one problem though. We have x^2 + 6x + 10, not x^2 + 6x + 9. However, we can put the (x + 3)^2 there, but we must add one to the (x + 3)^2, which ends up getting multiplied by the 1/2 to get a constant of one-half at the end. So, we have:

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

And that is vertex form. Let’s look at what is cool about it.

First off, both a’s happen to be equal. This is not coincidence, as we factored out the a in order to switch forms.

What I find really cool is those two terms we couldn’t control, h and k. In this case, they are -3 and 1 (the equation is a(x - h)^2 + k, not a(x + h)^2 + k, so h is -3). What does that have to do with anything? Take a look at the graph of this equation.

If you’ll notice, the vertex of this parabola (graph of a quadratic function) is in fact (-3, 1/2). In fact, you can actually graph a quadratic equation using only vertex form, just with this fact and a. I think that is cool that there is a format out there that can do that.

Bonus: Say that instead of doing a(x - h)^2 + k, you did (bx - h)^2 + k. b is unfortunately not the same as the other b, but much more useful. This form is messier, but there is a cool thing about it. In this case, the vertex is actually (bh, k), which is interesting.

What the a factor did was it told us the factor for the “vertical stretch/compression” of the parabola, or the number you multiply every single y value by. In the case before, every y value of plain (x + 3)^2 + 1 was multiplied by 1/2 to create the graph.

For this form, there is actually a “horizontal stretch/compression,” which isn’t common in quadratics. This form gives you a stretch/compression of 1/b. You might see it as just b, but it is usually written like this:

((1/b)x - h)^2 + k)

Either way, it is pretty cool.

Fibonacci Day: Fibonacci Magic Trick

2011-12-03T12:00:00.000-08:00

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.

Why √2 is not Rational. Or is it...

2011-11-26T12:00:00.000-08:00

Rational numbers are numbers that are the quotient of two integers. In other words, a number is rational if it is p/q, and p and q are integers (and q ≠ 0). An irrational number is a number that is not rational. For instance, numbers like pi, or e, the golden ratio, all of these are irrational. However, these numbers are hard to work with, so let's use √2.

What we will do is prove it is irrational. To do this, we will use a technique called proof by contradiction, by assuming the opposite and later getting into confusion. In this case, we will assume √2 is rational.

√2 = p/q

Assume p/q is in lowest terms, as we can write any rational number in lowest terms. In order to get rid of that square root sign and deal with easier numbers, let's square both sides.

(√2)^2 = (p/q)^2

2 = p^2/q^2

Now, let's multiply both sides by q^2 to get rid of the fraction.

q^2(2) = q^2(p^2/q^2)

2q^2 = p^2

p^2 = 2q^2

This means that p^2 must be even. In that case, p is even because an even squared is always even, an odd squared is always odd. So, we know p is even, or of the form 2a.

Let's plug 2a in for p.

(2a)^2 = 2q^2

4a^2 = 2q^2

2a^2 = q^2

q^2 = 2a^2

In this case, we are running into the same thing. Here, q^2 is even, meaning that q is even. So, we have p as even and q as even. However, we said that p/q is in lowest terms. This leads us to say that √2 is irrational.

I would never have thought that you could actually prove a number to be irrational. I'd assume you can with other numbers, such as π, e, the golden ratio, all of those guys. That is definitely one of my favorite proofs.

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

2011-11-19T12:00:00.000-08:00

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)

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!!!

2011-11-12T12:00:00.000-08:00

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.

Fibonacci Day: Even Fibonacci Numbers

2011-11-05T12:00:00.000-07:00

Today is a Fibonacci Day. It is the fifth of November and five is a Fibonacci number. Let's move on from the squares of Fibonacci numbers and just deal with plain old Fibonacci numbers. Here they are:

1 2 3 4 5 6 7 8 9 10 11 12
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

Which Fibonacci numbers are even? Well, we've got the 2, 8, 34, 144... Which numbers are those? They are F3, F6, F9, and F12. This time, the pattern is right in our faces!! They are all multiples of three! Why on earth is that?

Notice that the first three Fibonacci numbers follow the pattern of odd-odd-even. The next one is made up by adding the previous two. So, an odd plus an even is an odd. Then, the even plus the odd makes another odd. Then, the odd plus the odd makes an even. And we are back to where we started, odd-odd-even.

What's interesting is that every fourth Fibonacci number is a multiple of three. Every fifth Fibonacci number is a multiple of five. Every sixth Fibonacci number is a multiple of eight. Every seventh Fibonacci number is a multiple of thirteen, and the only multiples of thirteen! I think that is pretty cool!!

I'm not sure why that pattern continues, but if you know, please tell us. It would be pretty cool to see.

Bonus Proof: Since that was pretty short and simple, I'd like to show you one more little thing. You've probably seen that x^0 = 1, and wondered why. Why isn't it like zero or something? Well, let's take the number x/x. How do we simplify that?

Some people might say, it's a number over the same number, and everything over itself is one. That is absolutely correct. However, you might tackle it a little more algebraically, and realize that you have the same base raised to an exponent. You have x^1/x^1. We can use the law of exponents to subtract the 1 from the 1 to get 0, giving us x^0.

One way, we got one. The other way gave us x^0. Either way, we have that x^0 = 1. I thought that was pretty cool.

Cool Divisibility Stuff (There's a lot!!)

2011-10-29T12:00:00.000-07:00

This week, we are going to talk about divisibility, a pretty broad area of mathematics that we were lucky to be introduced to in fifth grade. I’ve looked into it more, and even use it for a lot of tricks in math.

There are a bunch of tricks for it, and I will explain them all and prove them to you. This post will almost be a lesson, but all of the rules are very cool! Let’s go through them all.

Divisibility by one: If the number has no decimal, it is divisible by one. For these rules, we are only working with decimals. To prove it, we look at the Closure Property, which states that any number multiplied by one will remain the same number. This means every integer is a multiple of one.

Divisibility by two: If the number ends in two, four, six, eight, or zero, it is divisible by two. The definition of even number is, “A natural number that is divisible by 2.” It is also defined as, “A whole number that has 0, 2, 4, 6, or 8 in the ones place.” Since these two things are the same, this means that this rule is valid.

Divisibility by three: This rule is probably one of the coolest, and most magical. If you add up the digits in the number and this gives you a multiple of three, this means that the number is a multiple of three. If the sum is too big for you to know if it has a three factor, go ahead and add up the digits again until you do know. I will prove this one along with the divisibility by nine rule, as it is very similar, and a number that is divisible by nine is also divisible by three.

Divisibility by four: Look at the number’s last two digits, and completely ignore the rest. If the number you are looking at is a multiple of four, then the whole number is a multiple of four. If you think about it, every number is 100x + y with y being the last two digits. Since we know 100x is a multiple of 4 (25x • 4 = 100x), all that leaves us with is y. This means that if y is a multiple of 4 also, then adding 100x won’t change this.

Divisibility by five: If the number ends in five or zero, it is a multiple of five. If you think about it, the digits from 0 to 9 end in either five or zero when multiplied by five. Since the last digit of a multiplication problem is the last digit of the last two digits multiplied together, we are left with all of the integral numbers ending in five or zero.

Divisibility by six: Since six is the product of three and two, we can just put both of these rules to use since they share nothing in common. If the number follows through with the divisibility by three rule (just adding to a multiple of three, no need for a multiple of six, 1 + 2 = 3 and 12 is a multiple of six), and is even, it is a multiple of six. Check the divisibility by two and three rules for more.

Divisibility by seven: This is also a really cool one, one my teacher didn’t even know. Take the last digit of the number and double it. Then, subtract that from the rest of the number. Keep repeating this until you have a one or two digit number (negatives are okay, you can just turn them positive). If the number you have is a multiple of seven, the original number is a multiple of seven. Let’s do one real quick, the number 224.

First, double the 4 to get 8. Subtract 8 from 22 to get 14. Since 14 is definitely a multiple of seven, the 224 is as well. 224 happens to be 32 • 7, so we did our math correctly. To prove this, we will use a little bit of Algebra. Our formula states that if 10r + x is a multiple of seven, then r – 2x is a multiple of seven. Let’s say the number is a multiple of seven. This means that r – 2x is a multiple of seven.

r – 2x = multiple of seven

Let’s try multiplying both sides by ten.

r – 2x = multiple of seven

10(r – 2x) = multiple of seven (and seventy)

10r – 20x = multiple of seven (and seventy)

Forget it is a multiple of seventy. We are not looking for that. However, let’s go in a different direction and add 21x to both sides. Since 21 is a multiple of seven, we still have a multiple of seven on the right-hand side of the equation. However, the left becomes:

10r – 20x = multiple of seven

10r – 20x + 21x = multiple of seven

10r + x = multiple of seven

And this brings us back to where we started. It is a little complicated, but I think it is pretty cool.

Divisibility by eight: Similar to divisibility by four, we will be looking at a block of digits on the right. However, note the fact that eight isn’t a multiple of one hundred. However, it’s a multiple of one thousand. So, we will look at the last three digits and see if it is a multiple of eight. Since this might be a little hard for you, as I don’t have my eights memorized that far, you can look at the hundreds place and see if it is odd or even. If it is even, this means that you can check the last two digits for divisibility by eight because 200 is a multiple of eight. If it is odd, check the last two digits for divisibility by four, but then make sure it is not a multiple of eight, or a multiple of eight plus four. Since 104 is a multiple of eight, this should work as well.

You can even take these principles to divisibility by any power of two. The power you are raising two to is the amount of digits you need to test at the end. So, for divisibility by 64, just check to see if the last six digits are divisible by 64 because 2^6 = 64.

Divisibility by nine: This is unquestionably the most magical of all of the proofs, and is used for so many things. It is the answer to numerous magic tricks, math tricks, and even provides loads of ways to check answers, including divisibility, digital roots, and mod sums. In fact, you can even cube root numbers based on this rule. To find divisibility by nine, you simply add up the digits of the number, and if that is a multiple of nine, the whole number is a multiple of nine. To prove it, let’s test the number 234. 2 + 3 + 4 = 9, so it is definitely a multiple. But why? Let’s write the number 234 in expanded notation, something you learn around third or fourth grade.

2 x 100 + 3 x 10 + 4 x 1

We can rewrite these to get:

2 x (99 + 1) + 3 x (9 + 1) + 4 x (0 + 1)

Let’s distribute the 2, 3, and 4 into the groups of numbers beside them.

(2 x 99) + 2 + (3 x 9) + 3 + (4 x 0) + 4

We can rearrange this to get:

[(2 x 99) + (3 x 9) + (4 x 0)] + 2 + 3 + 4

You might notice, but the sum inside of the brackets is definitely a multiple of nine, as it is being created by all multiples of nine. So, we can eliminate it to get the sum of all of the digits. Here, I don’t think the proof lives up to the magic in the actual trick, but it is pretty cool!

Divisibility by ten: To test divisibility by ten, all we need to do is see what the last digit is. If it is a zero, then the number is a multiple of ten. To prove it, we look at the Power of Zeros rule, which says that to multiply a number by a power of ten, just tack on that power amount of zeros. This means that a number multiplied by 10 has one zero at the end. You can elevate this to say that a multiple of 100 has two zeros at the end, a multiple of 1000 has three and so on.

Divisibility by eleven: To test divisibility by eleven, alternately subtract and add the digits and if you end with a multiple of eleven (it may be zero or negative), then your original number is a multiple of eleven. So for the number 1353, you would go 1 – 3 + 5 – 3 = -2 + 5 – 3 = 3 – 3 = 0. Since zero is a multiple of eleven, the full number is a multiple of eleven.

To prove it, we know that if you subtract eleven from a number constantly, the number still keeps its status as a multiple of eleven. Therefore, let’s put ten to its powers (creating the place values) and see what happens when you constantly subtract multiples of eleven.

10⁰ – (0 • 11) = 1

10¹ – (1 • 11) = -1

10² – (9 • 11) = 1

10³– (91 • 11) = -1

10⁴ – (909 • 11) = 1

10⁵ – (9091 • 11) = -1

10⁶ – (90909 • 11) = 1

and so on…

This convinces me enough, but I’m not sure of where to take it from there. If you want to add anything, please comment. It would be interesting for all of us.

Divisibility by twelve: Let’s take two distinct ones, divisibility by four and three. If the number is divisible by four and its digits add to a multiple of three, it is a multiple of twelve. Like the divisibility by six rule, this works as well. You can even take this principle to other numbers like this, like 14, 15, 18, 21, 22, 24, 28, and so on.

Divisibility by thirteen or greater (the “create a zero, kill a zero method”): Since this is a little more complicated, let’s learn it through an example. Is 2756 divisible by thirteen? First, we must look at the last digit of 2756, six. We need to find a multiple of thirteen that ends in six. If there is one, that means the number is not divisible, and you have learned a new property about that number. However, 13 x 2 = 26, so we have found one. We must subtract the 26 from 2756 to create a zero. So, 2756 - 26 = 2730. Now, we kill the zero, or just ignore it. Now, we create a zero by subtracting 13 (273 – 13 = 260). After killing it, we have 26. Since 26 is a multiple of 13, 273 and more importantly, 2756 are multiples of 13. You could even have subtracted 156 from the original number if you knew that it was 12 x 13. This would make it only take two steps instead of three.

Even though it is a lot in one dose, it is definitely worth practicing. These are really cool principles, and you could probably figure out a few tricks out of them. If you watch my Mathemagics show closely, I will use divisibility to perform one of the tricks, so you can investigate that as well.

How many cube roots of one are there?

2011-10-22T12:00:00.000-07:00

When getting into algebra, you will hear the term "real number" a lot, without an explanation as to why you can't just put "number." Since no one asks the question, it just slides by, and when the question is asked, the teacher responds, "You'll learn that in Algebra II," or something along those lines.

The answer to that question is that there is such thing as "imaginary numbers," or complex numbers, which are made up of a constant, a coefficient, and the letter i, which symbolizes the square root of negative one.

What is the square root of negative one? Some say negative one. Well, (-1) x (-1) = 1, so that is incorrect. People then turn around and say one. Well, 1 x 1 = 1, so that is incorrect. Then, they might try 1/2. Well, 1/2 x 1/2 = 1/4, so that is wrong. They will keep trying things until they give up.

What is the answer? If you think about it, a negative times a negative, or a negative squared, is a positive. A positive times a positive, or a positive squared, is a positive. So, you cannot square a real number and get a negative. So, mathematician Heron of Alexandria came up with the letter i, and began using that as the square root of -1. So, then, by the Multiplication Property of Square Roots, you can conclude that √(-9) would be 3i because you can break that into √(9)√(-1) = 3√(-1) = 3i. Rafael Bombelli built on his works, and make this concept a regular part of Algebra.

Now, let me introduce you to one more thing about imaginary numbers before I show you about the cube rooting. When you write these terms out, you write like 5 + 3i, which means 5 + √(-9), just like how you'd write a real square root in Algebra. The 5 + 3i would be known as a complex number, which is a number involving i. The conjugate of a complex number is to keep the same expression, but switch the operation separating them. For instance, the conjugate of 5 + 3i = 5 - 3i because we kept the same term, but switched the operation.

If you think about it, a complex number's conjugate is equal to the number because any number has two square roots, a positive one and a negative one. So by making the i term negative, we are just looking at the other root.

Now that we've gotten that out of the way, let's get to the good part! Let's take the complex number -1/2 + i/2√(3). It is the same thing, with a constant of -1/2 and coefficient of 1/2√(3). How about we cube it.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))(-1/2 + i/2√(3))

First, we'll square it. We can use FOIL for that. If you don't know, it stands for "First, outer, inner last." It's basically the distributive property made simpler for multiplying binomials.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) - i/4√(3) - 3/4
-1/2 - i/2√(3)

We ended up with the conjugate of before. That's interesting. Let's finish off by multiplying by the -1/2 + i/2√(3).

(-1/2 - i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) + i/4√(3) + 3/4
1 - i/4√(3) + i/4√(3)

What do we do with that? Well, there are i's in both terms, so we can combine them. However, look closer. They are opposites of each other, or the additive inverse of each other. What does that mean? The definition of additive inverses are two numbers in which when added together give you zero. So, these two confusing numbers simplify to zero!

1 - i/4√(3) + i/4√(3)
1 + 0
1

So, we are left with 1 as our answer! We did nothing wrong there. -1/2 + i/2√(3) is in fact the cube root of one, as well as its conjugate and of course, the integer one. Was this random? No! Mathematics is never random!

If you take the Cartesian Plane, and make the numbers going up the y-axis i, 2i, 3i, 4i, 5i, etc. and -i, -2i, -3i going down, you have the Imaginary Cartesian Plane. If you make a circle going through the points (1,0), (0, i), (-1, 0), and (0, -i), then you will have a unit circle. To find the 1st root of one, we of course start at (1, 0) and that is it. For the square root, or the second root, we would split the 360° of the circle in half to get 180°. So, we have the 1, and then we travel 180° to get -1, the other square root. For the fourth root, we could split 360 in fourths to get 90°, and at ninety degrees, all of the roots are found, 1, -1, i, and -i.

What about if we split in thirds, or 120°. Then, we end up at the points 1, -1/2 + i/2√(3), and -1/2 - i/2√(3). You can check that if you'd like. At the 72 degree marks, you will find the fifth roots, and the 60 degree marks give you the sixth roots.

If you know anything else about this, please tell us! Also, we will probably be taking more about imaginary numbers, so if you want me to show anything in particular, let me know.

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

2011-10-15T12:00:00.000-07:00

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.

Fibonacci Day: More Patterns in the Squares

2011-10-08T12:00:00.000-07:00

Today yet again is a Fibonacci Day! It is October 8th, and 8 is the sixth Fibonacci number. To keep the theme of last week, let's use the square Fibonacci numbers again. Here they are:

1 1 2 3 5 8 13 21 34
1 1 4 9 25 64 169 441 1156

Let's take each Fibonacci number and move one away from it. Now, we'll multiply those numbers and see how close we get to the square.

1) 0 x 1 = 0 = 1^2 - 1
2) 1 x 2 = 2 = 1^2 + 1
3) 1 x 3 = 3 = 2^2 - 1
4) 2 x 5 = 10 = 3^2 + 1
5) 3 x 8 = 24 = 5^2 - 1
6) 5 x 13 = 65 = 8^2 + 1
7) 8 x 21 = 168 = 13^2 - 1

And so on and so forth. Basically, Fn-1 x Fn+1 = Fn^2 ± 1, or even more accurate, Fn-1 x Fn+1 = Fn^2 + (-1)^n. Let's try it again, this time looking two away from the number. Keep in mind that 1 is the negative-first Fibonacci number.

1) 1 x 2 = 2 = 1^2 + 1
2) 0 x 5 = 0 = 1^2 - 1
3) 1 x 5 = 5 = 2^2 + 1
4) 1 x 8 = 8 = 3^2 - 1
5) 2 x 13 = 26 = 5^2 + 1
6) 3 x 21 = 63 = 8^2 - 1
7) 5 x 34 = 170 = 13^2 + 1

This time, we have pretty much the same pattern. Fn-2 x Fn+2 = Fn^2 ± 1, or Fn-2 x Fn+2 = Fn^2 - (-1)^n. How about we move three away. It's the same type of pattern, but a little different. Keep in mind that -1 is the negative-second Fibonacci number (since a Fibonacci number is the two numbers before it added together, than the zero comes from x + 1, or -1 + 1).

1) -1 x 3 = -3 = 1^2 - 4
2) 1 x 5 = 5 = 1^2 + 4
3) 0 x 8 = 0 = 2^2 - 4
4) 1 x 13 = 13 = 3^2 + 4
5) 1 x 21 = 21 = 5^2 - 4
6) 2 x 34 = 68 = 8^2 + 4
7) 3 x 55 = 165 = 13^2 - 4

We have the same idea. We are stuck with a four, giving us the pattern of Fn-3 x Fn+3 = Fn^2 + 4(-1)^n. Let's look at our neighbors four away and see if we can see the pattern better. What do you think the negative-third Fibonacci number is? If you got two, then good job.

1) 2 x 5 = 10 = 1^2 + 9
2) -1 x 8 = -8 = 1^2 - 9
3) 1 x 13 = 13 = 2^2 + 9
4) 0 x 21 = 0 = 3^2 - 9
5) 1 x 34 = 34 = 5^2 + 9
6) 1 x 55 = 55 = 8^2 - 9
7) 2 x 89 = 178 = 13^2 + 9

Same idea again. We have Fn-4 x Fn+4 = Fn^2 - 9(-1)^n. However, the four and nine aren't there randomly. Let's look these differences closer.

1, 1, 4, 9

Recognize them? They are the squares of the Fibonacci numbers again! If you go five away, it is the square of the fifth Fibonacci number, six away is the square of the sixth Fibonacci number, one hundred away is the square of the hundredth Fibonacci number. Basically, a general formula is Fn-a x Fn+a = Fn^2 ± (Fa^2)(-1)^n. Or, you can use the below formula to be even more accurate:

Fn-a x Fn+a = Fn^2 - ((-1)^a)(Fa^2)((-1)^n)

I have no clue why this works, but please put up a proof if you know it. This is one of the coolest things about Fibonacci numbers!

Fibonacci Day: Adding the Squares

2011-10-01T12:00:00.000-07:00

Today is a Fibonacci Day! It is the first, and one is a Fibonacci number. One is in fact two Fibonacci numbers. In the last post, you learned how to square numbers that end in five. To keep this squaring theme, how about we square the Fibonacci numbers.

1, 1, 2, 3, 5, 8, 13, 21, 34...
1, 1, 4, 9, 25, 64, 169, 441, 1156...

We've been doing lots of adding Fibonacci numbers. Let's finish it off by adding the square Fibonacci numbers.

1 = 1
1 + 1 = 2
1 + 1 + 4 = 6
1 + 1 + 4 + 9 = 15
1 + 1 + 4 + 9 + 25 = 40
1 + 1 + 4 + 9 + 25 + 64 = 104

Do you see a pattern? It is a little hard to find, but definitely present. Look at this:

1 = 1 = 1 x 1

1 + 1 = 2 = 1 x 2

1 + 1 + 4 = 6 = 2 x 3

1 + 1 + 4 + 9 = 15 = 3 x 5

1 + 1 + 4 + 9 + 25 = 40 = 5 x 8

1 + 1 + 4 + 9 + 25 + 64 = 104 = 8 x 13

They are the product of the two consecutive Fibonacci numbers! Why on earth would that be? I had recently looked for one on the internet, and found an amazing geometric proof for it.

Have you ever heard of the golden rectangle, or the golden ratio? We touched on the golden ratio when I gave you the explicit formula for Fibonacci numbers (the golden ratio is the same as the greek letter fi). The golden rectangle is a rectangle of which the ratio of the length and width is the golden ratio. Something else whose ratio is the golden ratio is Fibonacci numbers! So, the side lengths of the rectangle are consecutive Fibonacci numbers!

Since we are dealing with squares of Fibonacci numbers, let's make some squares.

We've just taken these squares and organized them in a fashion that makes the side lengths two Fibonacci numbers. We went up to 34 squared, so let's see what the side lengths are.

They are 34 and 55. So, to figure out the area of the whole thing, you can add up the areas of all the squares, or just multiply the 34 by 55. And because of the way it is laid out, you can do it with any Fibonacci numbers! I think that is really cool!

Bonus Pattern: How about we add the squares of consecutive Fibonacci numbers.

1 + 1 = 2

1 + 4 = 5

4 + 9 = 13

9 + 25 = 34

25 + 64 = 89

64 + 169 = 233

The sums of the consecutive square Fibonacci numbers is in fact a Fibonacci number. I don't know a proof for this, but please tell me if you find one!

Check out my Mathemagics Performance, and even learn a trick!!!

2011-09-24T12:00:00.000-07:00

I don't know if you have seen my performances before, so today is your opportunity to see it. Basically, I spend the show doing very big multiplication and squaring problems, create magic squares, tell people the day of the week they were born on, and more. Here is a compilation of some of my performances.

If you liked it, you can see me live at the Chicago Toy and Game Fair in November, and many other possible shows. These are listed on my website, www.EthanMath.com.

In this video, you saw me square numbers. This is a method like the close-together method, but even easier than squaring numbers, and even the things I've taught you all along is squaring a two-digit number ending in five. A crowd of fifth graders could understand it, so hopefully, you guys are smarter than those fifth graders!!

Let's take the number 35. In order to square it, you need to remember two things:

1) The last two digits are always 25.
2) The start of the number is the first digit times the first digit plus one.

So, first, we take the first digit, or 3. Three times four (3 + 1) is equal to 12. Since the last two digits are always 25, 35^2 is 1225.

Let's look at, say 85^2. 8 x 9 = 72, so the answer is 7225. Very simple, right.

Let me make it a tad harder on you. Try 145^2. All we need to do is think of it as a two-digit number, with fourteen as the first digit and five as the last one. So, 14 x 15 isn't too bad, as you remember from the first post. If you try it yourself, you will get the number 210. Since it always ends in 25, the answer is 21025.

With very little practice, you will be able to square two-digit numbers ending in five. If you practice the close-together method I taught on August 20, you will be able to square three-digit numbers ending in five. Have fun!!

August Problem of the Week Answers:

Easy:

b = 35
Explicit Formula: n - 1
Recursive Formula: an-1 + 1
g = 34
a = 908
p = 5%

Hard:

z = 42.5
a = 1/2
b = 1/2
c = 1
y = 55
x = 10

What does .9999999999999... really mean?

2011-09-17T12:00:00.000-07:00

Haven't we all heard people say that there is a 99.999 "repeating" percent chance that something will happen. They think that this means they are almost sure it will happen. Though I don't want to make things complicated, this is actually equal to 100 percent.

To make things more simple, let's look at if 0.9999... = 1. First off, realize that 1/3 x 3 = 1, but .33333... x 3 = .99999... Same with 1/7, or other repeating fractions.

However, this won't convince people. So, I like to show them my favorite way to look at it, the algebraic proof, the one that proves this fact, or all of algebra and other mathematics incorrect. Say that .9999... is equal to S.

S = .9999...

Try multiplying that all by ten.

10S = 9.9999...
   S = .9999...

What if we subtract both of these equations from each other.

10S = 9.9999...
   -S = .9999...
  9S = 9
   S = 1

By dividing both sides by nine, it shows that S is both of these numbers.

This is another thing that is absolutely shocking. Before I saw this, I always thought that it was too close to call, but not exactly one. This is probably one of my favorite proofs in number theory, and all of mathematics.

Another Probability Paradox: What's your birthday?

2011-09-10T12:00:00.000-07:00

As I said about a month and a half ago, probability is a topic that people have difficulty understanding. I showed you the Monty Hall Paradox, where the average person is almost 17% off on the odds! 17% makes a big difference in what you should do in that scenario! Let's look at another thing that completely fools people, the Birthday Paradox.

Say you walk into a room of 23 people. What do you think the odds are that two of the people have the same birthday? Maybe like 23 in 365 because there are 365 days in a year, being about 6.3%. Would you be surprised if the true answer was over 50%?

The mistake people make is they try to determine the odds that someone has the same birthday as them. This is not correct, as there is no specification in the problem as to which two people it is. If you were to calculate those odds, it would be around 0.14%, which is nowhere close to the true odds.

In the Monty Hall Problem, I showed you how we know this. For this problem, we don't need much proof, because we can figure out the odds. Not of this question, but of the opposite question.

First off, what if we want to make a list of how many possibilities there are for 23 people. There are 365 for the first person, 365 for the second, and so on. Basically, there are a total of about 85 octodecillion (8.5 x 10^58), or 365^23. Basically, a list for n people is 365^n.

Now let's cover the rest of it. How many of these different possibilities have all different birthdays? Well, the first person's birthday could be any of 365 days. Since the second must be different, it only can choose between 364 days. The third only has 363, all the way up to the 23rd, who has 343 different possibilities. For n people, this is equal to:

365 x 364 x 363 x ... x (367 - n) x (366 - n)

You can very easily shorten this expression. On your calculator, there may be a button that is an x with an exclamation point after it. All this does is takes the number you type in and multiply it by every single whole number before it. For instance, 6! is 720 because 6 x 5 x 4 x 3 x 2 x 1 = 720. So, this expression is equal to:

365!/(365-n)!

And this is out of a list of 365^n possibilities. Therefore, the total possibilities there are for no one having the same birthday is:

365!/(365^n)(365-n)!

If you subtract this decimal from one, you will know the probability that two people have the same birthday. For 23 people, we 365!/(365^n)(365-n)!]

This is about a 50.7% chance. For 30 people, we are already at 70%. 50 people is already at a 97% chance. 100 people is like a 99.99996% chance that two people will have the same birthday.

Fibonacci Day: Adding the Odds

2011-09-03T12:00:00.000-07:00

Today is another Fibonacci day. It is the 3rd, which is a Fibonacci number. We've looked at a bunch of addition patterns in the Fibonacci numbers, like adding them, or the even ones. What about the odds? How about we look.

1 = 1
1 + 2 = 3
1 + 2 + 5 = 8
1 + 2 + 5 + 13 = 21

See the pattern? We are getting the Fibonacci numbers! Why? Let's look at each one as the previous two.

1 + (1 + 1) + (2 + 3) + (5 + 8)

You are basically adding the Fibonacci numbers up, and then adding one! that's a Fibonacci number minus one plus one, or a plain Fibonacci number.

Bonus: We've done quite a bit with explicit formulas lately. I think now is the time to mention the explicit formula for Fibonacci numbers. For the nth Fibonacci number, you do:
_
[1/(5^.5)](φ^n - φ^n)

If you don't know, the Greek letter fi means the golden ratio, or 1.618... Fi with a bar on top is what you get with a small change in the golden ratio's formula, -0.618...

This is definitely complicated, but I couldn't believe that worked. I checked many numbers just to convince myself it worked!!

Do the Primes go to Infinity and Beyond?!!

2011-08-27T12:00:00.000-07:00

We've been doing a lot of work with patterns, or more formally called sequences or infinite series. Infinite series are a little different though, as they are guaranteed to go on forever without stopping. For instance, the sequence with formula n^2, or 1, 4, 9, 16..., is an infinite series. However, one like √-n + 3 is not infinite, as it will go √2, 1, 0, and then will be hitting complex numbers, which are not valid for these sequences.

Most patterns, it is pretty obvious, and would never be on a test, or even a valid question for a teacher to ask. However, what about the prime numbers. If you don't know, primes are numbers with only two factors, one and itself. So, 5 is prime, because its factors are 1 and 5. However, 6 is not prime, because it has more than two factors, namely 1, 2, 3, and 6. That means that 6 is composite, which is having three or more factors. Numbers such as one are known as universal, as they have only one factor.

Anyways, are prime numbers infinite? This is definitely a valid question, and answerable too! Can you create a list of all prime numbers on it? How about you try to. I'll bet you can't.

Say someone pops up and says,"I have made a list with all of the prime numbers that exist on it." The list would be much longer if someone did say that, but I made a small list below:

2
3
5
7
11

Okay. Let's multiply all of these numbers you've found together. 2 x 3 x 5 x 7 x 11 = 2310. Great. According to you, this is the product of all of the prime numbers out there. Try adding one. Now, we have 2311, which is a multiple of no prime numbers. Since every number is composed of primes, or is prime, this is not possible. That means the number is either prime, as 2311 happens to be, or could be a multiple of another prime that is not present on the list.

Is that all of the primes? No! We can do that process forever, and always find a prime that is missing. This is not a formula to generate prime numbers, as the first ten primes multiplied together is 6,469,693,230 which if you add one gives you 6,469,693,231, which you have no clue if it is prime or not! You guys can figure that one out. However, it is a cool proof that answers a question that definitely gets you thinking!

Multiplying Giant Two-Digit Numbers Together with No Effort!!!

2011-08-20T12:00:00.000-07:00

In the first week, we learned how to multiply two numbers in the teens together instantly, as if you memorized the answer. To the average person, this is the only way possible to get the answer so quickly! Now, you can take that up a notch, and multiply together the biggest two digit numbers. Say someone says 97 x 94. You can say, "That's easy! It's 9118!"

We will take it step by step here. For the first two digits, you take the bigger number and see how far away it is from 100. In this case, it is three away. Then, you subtract that from the other number. 94 - 3 = 91. There's your first two digits.

For the last two digits, take how far both numbers are from 100. 97 is 3 away and 94 is 6 away. The answer is precisely 3 x 6 = 18. Put them together and you have 9118.

Let's try another one, 95 x 89. 95 is 5 away from 100, and 89 - 5 = 84. Then, 89 is 11 away from 100, so 11 x 5 = 55. Then, we put them together to get 8455.

You are probably wondering why this works, for teens or nineties. Let's start with the teens. Pretend that the problem is (z + a)(z + b) with z being 10. We will leave it as z for the moment.

(z + a)(z + b) = z(z + a + b) + ab

If you factor it out, the z(z + a + b) becomes z^2 + za + zb.

(z + a)(z + b) = z^2 + za + zb + ab

If you FOIL out the (z + a)(z + b), you get:

z^2 + zb + za + ab = z^2 + za + zb + ab

This shows that they are equal. If you think about it, this is what we are doing. Take 17 x 16.

(10 + 7)(10 + 6) = 10(10 + 7 + 6) + (7)(6)

This is what you are actually doing. What about for 89 x 95.

(100 - 5)(100 - 11) = 100(100 - 5 - 11) + (-5)(-11)

I got commented about the fact that 20 wouldn't work for the teen method I described in the first post. Like 20 x 18 wouldn't work. However, this formula directs us to do it as so.

(10 + 10)(10 + 8) = 10(10 + 10 + 8) + (10)(8)

This will give us the 360 as promised. If you can do 2 x 1 multiplication problems in your head, you might want to try things like 32 x 37 with 30 as your z, or 68 x 66 with 70 as your z. If you get really good at that, you could even try doing 448 x 442 with 400 as your z, which makes you add 48 x 42, using 40 or 50 as your z. This would be difficult, but you would get 198016 as an answer.

Problem of the Week Solutions (from July):

Easy:
b = 16
z = 1
n = 42
p = 116
odds = 58%

Hard:
s = 53.1 degrees
t = 36.9 degrees
a = 1
b = 8
c = 0
h = -4
k = -16
area = 50.3 sq. cm

The Problem of the Week Day 5: Week of 8/14 - 8/20

2011-08-19T12:00:00.000-07:00

Today, we will finish up August's problems. Since the easy problem usually has an order of operations problem and a probability problem, we will have both. Since we did deal with chess this week, the probability problem requires knowledge of chess. If you are not familiar with chess, please do the other problem, because I don't want you to get the wrong answer because of expertise in an area besides mathematics. However, the hard problem has only one way to find the answer, which is something I have not introduced yet.

Easy Problem:

Probability: If a chess player knows nothing about chess, and makes a completely random first move, what are the chances he will do d4 as his first move?

p = ___

Order of Operations: p = (25b + g - a - 3e)/3

p = ___

Hard Problem: Today, I will introduce a great tool in Algebra, the Quadratic Formula. The Quadratic Formula states that if ax^2 + bx + c = 0, then x = (-b ± √(b^2 - 4ac))/2a. So, if x^2 - 5x + 6 = 0, then to find x, you would do:

(-(-5) ± √(5^2 - 4(1)(6)))/2(1)
(5 ± √(25 - 24))/2
(5 ± 1)/2
(5 + 1)/2 OR (5 - 1)/2
6/2 OR 4/2
3 OR 2

x = 3
x = 2

1) I made a couple of errors on Monday's problem. Please complete these simple calculations to have the correct z. We will call this number y.

y = 4z/5 + (4)(5) + 1

y = ___

2) If Ax = y, what does x equal? Use the explicit formula you created with the quadratic formula to achieve the answer.

Tip: There will be two possibilities. Choose the one that is reasonable. For instance, if you had 3 and -5, 3 would be the correct answer because you cannot cut a pizza with -5 straight lines.

x = ___

Also, let's call your false answer f.

f = ___

The Problem of the Week Day 4: Week of 8/14 - 8/20

2011-08-18T12:00:00.000-07:00

Today, the easy problem will move on and work on some geometry. For the hard problem, we will still be cracking away at the pizza problem!

Easy Problem: You probably know that to find the area of a circle, you square the radius and multiply it by π (3.14...). However, that is a little too easy. So, we will try finding the area of a quarter circle. In order to do that, we will do (πr^2)/4, or the radius squared times pi divided by four. Basically, you've found the area of the circle, and divided it by four.

The number of games it takes for b players will be called g.

What is the area of a quarter circle with radius g? Round to the nearest whole number this time.

a = ___

Hard Problem: You should remember the process behind solving a system from the last few months. If you don't, you will just create a variable with the same coefficients, and subtract the two equations. Continue this until you have isolated a variable. Then, you will substitute to find the rest.

Solve yesterday's system from the pizza problem. What is the explicit formula for this sequence?

Hint: It is quadratic (polynomial of degree two).

If you want, try finding the recursive formula. There is a systematic way to do it, which I'll bet you can figure out! You are solving a system! Remember, the recursive formula is An-1 + d with d being the difference.

The Problem of the Week Day 3: Week of 8/14 - 8/20

2011-08-17T12:00:00.000-07:00

Today, we will finish up on the chess tournament problem. You should have already noticed a pattern.

Easy Problem: There are two types of formulas you can have in a sequence. One of which is the explicit formula, which is a formula based on the nth term. For instance, the explicit formula for the sequence 2, 4, 6, 8, 10, ... would be An = 2n. If you wanted the 6th tern, you would plug 6 in for n to get 2(6) = 12.

The other type of formula is the recursive formula, which is based on the previous term. In the even number sequence, the recursive formula would be An = An-1 + 2 because it is the previous term plus two.

1) What is the explicit formula for the chess tournament problem?

2) What is the recursive formula for the chess tournament problem?

3) If there were b players in the tournament, how many games would it take to find a winner fairly?

Hint: Use the explicit formula for number three!

Hard Problem: At CTY, we used various strategies to determine explicit formulas. However, I tend to lean towards the method I taught last month, with the systems. Just to remind you how to find the system, you must first find your common differences. For the sequence 2, 4, 6, 8, 10, ..., the differences are in the first row. Therefore, you are dealing with a first-degree, or linear, equation. So, we take our base equation for this: An = mn + b. Then, plug in values for n and An to create the system. For instance, you would first plug 1 in for n and 2 in for An, and have the first equation, m + b = 2. Then, you would create a second one, 2m + b = 4, to get our constants, m = 2 and b = 0. This makes the explicit formula An = 2n + 0 which becomes An = 2n.

1) Find common differences in the Pizza Problem.

2) Create a system of equations.

If you want to get ahead, try solving the system, or even try finding a recursive formula.

The Problem of the Week Day 2: Week of 8/14 - 8/20

2011-08-16T12:00:00.000-07:00

Today, we will do start our main problems! We will mainly be generating values to work with today.

Easy Problem: If there is a chess tournament with n people in it, how many games must they play to get a winner?

For example, if there are seven players. One player will get a bye (sit out for the round and automatically move on) while the other six play. There will be three winners plus the bye makes four. Then, they will play two games to get the two finalists who will play for the win. So, in the first round, there were three games, then two, then one, giving you a six game tournament.

Find data for the values from 1 - 6.

n	1	2	3	4	5	6	7
An							6

Hard Problem: If you cut a pizza with n straight cuts, what is the maximum pieces of pizza you can get?

If you were to use one straight cut, you would just be able to split the pizza in half, which would make two pieces of pizza.

Find data from 2 - 5 or 6.

n	1	2	3	4	5	6	7
An	2

Hint: The pieces don't have to be equal in size, and shouldn't be equal.

The Problem of the Week Day 1: Week of 8/14 - 8/20

2011-08-15T12:00:00.000-07:00

Since I just recently came back from CTY, the problems will mainly be a problem we discussed in class. The easy one will be a problem about a chess tournament, and the hard will be about slicing pizzas. However, we are going to start off with some triangles as always!

Easy Problem: In the last two problems, we worked with the famous Pythagorean Theorem, developed to find the missing side of a right triangle when given two. Just to remind you, the shortest side is labeled a, the medium labeled b, and the longest labeled c. The formula states that a^2 + b^2 = c^2.

If you have a right triangle with a equalling 26.25 and with c equalling 43.75, what is the value of b? Remember to round to the nearest tenth!

b = ___

You won't need b's value until Wednesday, so hold onto it after you find it.

Hard Problem: Last month, we used the button on the calculator that turns a sine into its angle. Let me review how it works.

"You might not have it on your average calculator, but if you hit the 2nd button on a scientific calculator or iPhone calculator, you will get a button where the sine function has a little -1 above it, in the place of an exponent. That button takes the sine of an angle, and turns it into the angle. So, you could divide the side opposite to an angle by c and get the sine, and then hit that button to retrieve your angle."

If a right triangle's sides are lengths 27, 29.5, and 40, what is the measurement of the angle opposite to side a? This angle will be referred to later on as z.

z = ___

You will not need this value until Friday.

Additional Challenge: At camp, we also had a puzzle called region revenge. It was pretty difficult, but I'd like to share it with you. I will put up the answer in two months.

The problem is basically to create a vertex and see how many shaded areas are in the circle. So, for one vertex, there is one shaded region. Now, make two vertexes and connect them with a line. That makes two shaded areas. Then, you make three, and connect every vertex to one another, making a triangle. There, there would be four shaded areas. Keep doing this for five to seven circles, and look for differences. Can you find an explicit formula? It is not what you think it is.

Hint: The formula is a quartic equation (equation of degree four).

Fibonacci Day: Adding the Evens

2011-08-13T12:00:00.000-07:00

If you look closely, you'll find that today is in fact a Fibonacci day. 13 is the seventh Fibonacci number, so we're in for a treat!

Last month, we added up the Fibonacci numbers. This time, let's add the Fibonacci numbers in the even positions. We'll list some out.

1 = 1
1 + 3 = 4
1 + 3 + 8 = 12
1 + 3 + 8 + 21 = 33

See the pattern? Same thing as last time! Just the odd spots minus one!

1 = 1 = 2 - 1
1 + 3 = 4 = 5 - 1
1 + 3 + 8 = 12 = 13 - 1
1 + 3 + 8 + 21 = 33 = 34 - 1

Why does this work? Let's split each number into the previous two Fibonacci numbers.

1 + 3 + 8 + 21 = 33 = 34 - 1

1 + (1 + 2) + (3 + 5) + (8 + 13) = 34 - 1

Because of the associative property, we can eliminate the parentheses, giving us what we did last time, adding all the Fibonacci numbers. That would always give us a Fibonacci number minus one as well.

If you know something cool about Fibonacci numbers, please let me know!

CTY Challenge: Calculating pi like in the olden days!

2011-08-06T12:00:00.000-07:00

This is the last week of CTY camp. Of course, a few activities deal with π because, who doesn't love pi? We did two activities that were attempts on calculating pi, with very little to use. Of course, we did the old make a circle and divide the circumference by the diameter. However, we also did a variation on the Buffon's Needle experiment.

The problem goes like this: take a ruler and draw a bunch of parallel vertical lines that are two inches apart and a foot or two long. Then, take a two inch needle and toss it onto the grid and determine if it crosses the gridlines or not. What do you think the probability is that the needle will cross the line?

If you do the work, you should get a number around 63.66%. However, try plugging in these values into this equation:

2t/c =

t = total number of tosses
c = total amount of times the needle crosses the line

What do you get? Pi, exactly. Our whole class tried it, and with over 1000 tosses, our class average is 3.35. Pretty good, right?

Our instructor never went over the proof in class, though it involves geometry, calculus, and more advanced statistics. I think it has to do with the center point of the needle, and where it is in relation to the gridlines. If you have a proof, please post it!

Additional Puzzle: A census taker walks up to a house, and records the house number. Then, he knocks on the door and a man answers the door. The census taker asks if anybody else lives with him. The man responds that he lives with his three children. The census taker then asks for their ages. The man responds, "Their ages add up to the number on the door and their product is 36." The census taker then says, "I need one more clue. Is the youngest child a twin?" The man says that the youngest child is not a twin. What are the kids' ages, and what is the number on the door?

Hint: Find all the possible combinations of three numbers multiplied together to get 36 (including ones where the smaller numbers are equal).

More Patterns at CTY: All in one triangle!!

2011-07-30T12:00:00.000-07:00

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?

Patterns and Puzzles at CTY

2011-07-23T12:00:00.000-07:00

For the next few weeks, I will be at Johns Hopkins Center for Talented Youth Program studying something called "Inductive and Deductive Reasoning." I will be posting anything I learn during the week that would appeal to you.

We learned some really cool patterns that you guys will definitely like. The first of which involves just ones. Take 1 x 1.

1 x 1 = 1

Now, try 11 x 11

11 x 11 = 121

How about 111 x 111

111 x 111 = 12321

Let's try them all the way through to 9 ones. Pull out your calculators and you should see that it does. What about with ten ones? Sure enough, the pattern does continue, but in disguise. Try opening up a spreadsheet (unless you own a 19+ digit calculator) and type it in. You should get 1234567900987654321. This is because the 10 couldn't fit in that spot, so the zero dropped in and the one carried over to the nine which carried to the eight, giving you this answer. And this pattern I believe continues (if you find a proof, please inform me of it or post it for us) forever just by carrying ones and what not.

Also, let's try taking these sequences of 123456... How about we go on and stop. Then, we multiply by eight and add the number we stopped at. Let's see:

1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654

This pattern continues all the way up to 123456789. I'm not sure what happens after that, but it's still really cool. If you have a proof of this, please post it!

Bonus: In class, we've done a few puzzles. I would like to share one I really liked with you.

You are by a river with only a nine quart bucket and a four quart bucket (ONLY that, no bucket to dump it in at the end), and need to bring exactly six quarts of water back from the river. Following these guidelines, how would you get the water?

Email me the answer you get and I will tell you if it's right. If you want the solution, I will tell you, but don't spoil it for the others!!

Problem of the Week solution (from June):

Easy Problem:
c = 5
z = 3
n = 6
y = 1
odds = 71.4 %

Hard Problem:
a = 31.2 in
b = 24.1 in
x = 1
y = 3
m = -3
r = 6
n = -1
g = 5/3
q = 5.2
p = 9.5
z = 12.3
h = 8.2
area = 89.38 or 89.4 cm^2

The Problem of the Week Day 5: Week of 7/17 - 7/23

2011-07-22T12:00:00.000-07:00

Today, we complete the problem of the week! For the easy equation, you need every variable, while the hard one requires just yesterday's answer.

Easy Problem: Today, it is another experimental probability question. Just to review, you take the amount of things you are looking for and putting it over the amount of things total. Then, convert to hundredths and make a percentage.

If I have p - b Jolly Ranchers with n being blue raspberry and p ÷ 2 being watermelon. If I pull z jolly rancher(s), what are the odds that it will be watermelon? Express answer in a percent.

Hard Problem: To find the area of an ellipse (better known as an oval), you must determine the shortest and longest radius, and find their product. Then, multiply that by π to get your area. Here, round to the nearest tenth.

If your radii are -h cm and -k cm, what is the area of your ellipse? Your answer should be in square centimeters.

The Problem of the Week Day 4: Week of 7/17 - 7/23

2011-07-21T12:00:00.000-07:00

We have completed our sequences! For the easy one, we will just be using n and b, and the hard one only requires your equation. For the hard problem, the equation should be quadratic, or you made a mistake. If it is linear, go over Tuesday's work and find your mistake. Then, catch up so you can do today's work.

Easy Problem: To find the perimeter of a polygon, add up its sides. If you have a rectangle with sides 5, 5, 13, and 13, add them all up to get a perimeter of 36. On a rectangle, you can determine the perimeter with the formula P = 2b + 2h with b being the base and h being the height of the rectangle.

If you have a rectangle with n as the base and b as the height, what is the perimeter of the rectangle?

p = ___

Hard Problem: Quadratic equations have two forms they can be written in. The one we used is called "standard form," with the equation in the form ax^2 + bx + c. The other form is called "vertex form," being in the form a(x - h) + k. This is called vertex form because the vertex, or turning point, of the parabola (the graph of a quadratic equation, looking somewhat like the letter U) is (h, k).

To go from standard form to vertex form, you do something called "completing the square." Say you had the equation y = 2x^2 + 4x - 6. First, you factor your a term out of the equation to get y = 2(x^2 + 2x - 3). Then, you complete the term by dividing your x coefficient (not x^2) by two and squaring it. 2/2 = 1 which squared is 1. That means that x^2 + 2x + 1 is a perfect square trinomial. Therefore, we need to add one to the x^2 + 2x, which means we also have to subtract one to even it out. This gives us y = 2(x^2 + 2x + 1 - 1 - 3). Now, we make the x^2 + 2x + 1 a square, with the h term being the square root of the number you added and subtracted. This gives us y = 2((x + 1)^2) - 1 - 3). Then, we combine the -1 and -3 to get y = 2((x + 1)^2 - 4). By distributing the 2 over to the -4, we get our equation to y = 2(x + 1)^2 - 8.

Tip: Vertex form is y = a(x - h) + k. It is NOT x + h.

1) Put your equation from yesterday into vertex form. It will take less time to do.

2) Find the vertex of the equation.

h = ___
k = ___

The Problem of the Week Day 3: Week of 7/17 - 7/23

2011-07-20T12:00:00.000-07:00

Today, everybody get into sequence mode! We will have a lot of fun with these problems! Good luck.

Easy Problem: In order to find the next number in a simple sequence, look and see what you are adding/subtracting to get to the next number. If you find that they are the same, then you should be able to find the next number in the sequence. If they are not, see if you are multiplying or dividing by something to reach the next number. If so, you can also reach the next number in the sequence. That is called a geometrical sequence.

1) Plug z and b into this sequence:

z, b, 29, ___, 55, 68, 81, 94 ...

2) Look for a pattern in this sequence. It shouldn't be hard to find.

3) Determine what the number in the blank is by using this pattern. We will call that number n.

n = ___

Hard Problem: Here, we will solve our system and create an equation. Last month, we learned how to solve systems. If you remember that, you can eliminate the b's or c's by using "The Elimination Method." Just to review, if you have two variables that are the same and have the same coefficient (number to the left of the variable that is multiplied by the variable), you can subtract both equations from each other to create a different equation with different variables. If you have an equation with three unknowns (a, b, and c), create two equations with this method, and then solve for that system. Then, plug those answers into an equation from the original system to get your third. Don't forget, a three-unknowns system requires three equations. If you only had two, create a third one.

1) Solve the system created from yesterday. If you have two variables, they should be m and b, and three variables is a, b, and c.

Linear Answer (if that was your system):

m = ___

b = ___

Quadratic Answer (if that was your system):

a = ___

b = ___

c = ___

2) Plug these answers into y = mx + b or y = ax^2 + bx + c to find your equation. This equation should determine any number in the system.

3) Just for fun, you should try figuring out the next number in the sequence, or maybe the spot for your favorite number. You can even find a fractional, negative, or imaginary value in the sequence. Or, you can see what spot is your favorite number by solving a two-step equation or using completing the square, factoring or the quadratic formula. However, this is optional.

The Problem of the Week Day 2: Week of 7/17 - 7/23

2011-07-19T12:00:00.000-07:00

In the next few days, we will be looking at a topic called "Sequences and Pattern Recognition," which is basically finding the next number in a pattern. In first grade, you make the little bumps with a plus two on top or something to get a feel for simple patterns. We are going to take it to the next level, and look at a quadratic pattern which will use systems of linear equations to find. These are one of my favorite parts of Algebra.

Easy Problem: Since Tuesday, we have been solving equations, we will keep that going. Tomorrow, we will look at a sequence, but today, we'll just keep it simple. In our equation, we will have two steps after you plug in b and simplify. First, you will have to get rid of some addition at the end by subtracting from both sides. Then, you will divide on both sides to reach your answer. Good luck!

Plug in b and solve for z: 100 = 6bz + 4

z = ___

Hard Problem: When given a sequence, it is essential that you create an equation that has you put in the spot that you are looking for and have it equal to the number in that spot. For instance, the sequence 3, 12, 21, 30, ... will have the equation n = 9x - 6 with x being the spot you put it in and n being the value in that spot. In order to figure out the equation, you need to look for common differences (for arithmetic sequences). In that one, you would have:

3 12 21 30
\/ \/ \/
9 9 9

In that case, we had the same differences at our first step. That means our equation is in the form y = mx+b. Hence, we create a system where we are solving for m and b by plugging in 1 for x and 3 for y in the first one. 2 for x and 12 for y is the second, and so on. You'll notice these equations are very easy to do elimination in because terms are already isolated.

If you don't get common differences, try finding the common differences of the common differences and see if those are the same. If they are, plug values into y = ax^2 + bx + c and solve for a, b, and c. This will require three equations.

1) Find the value for p and q.

s + 0.9/6 = p
t - 3.9 = q

2) Find the common differences in this sequence: p, 20, q, 48, ...

3) Create a system that could be used to find the equation for this sequence.

If you want to put yourself one step ahead, try to solve the system and determine what the equation is. Since we went over systems last month, you should be able to figure it out. Just remember to eliminate variables with no coefficient. Tomorrow, everybody is doing sequences! It's going to be fun!!

The Problem of the Week Day 1: Week of 7/17 - 7/23

2011-07-18T12:00:00.000-07:00

The problem of the week is back for its second time! Make sure you did last month's problem because the answers are going up this Saturday. Don't forget to round to the nearest tenth unless told otherwise.

Easy Problem: Last month, we learned about the Pythagorean Theorem and how to figure out the missing side of a right triangle. Just to refresh your memory, the square of the longest side: c, equals the sum of the squares of the two shorter sides, a and b. So, a^2 + b^2 = c^2.

If you were missing a and b, it is just another algebra equation, but the inverse operation to squaring is square rooting.

If you have a right triangle with a =12 and c = 20, what does b equal?

b = ___

Hard Problem: Last month, we learned about the sine function. Now, we will look at another function that you might not have on your average calculator, but if you hit the 2nd button on a scientific calculator or iPhone calculator, you will get a button where the sine function has a little -1 above it, in the place of an exponent. That button takes the sine of an angle, and turns it into the angle. So, you could divide the side opposite to an angle by c and get the sine, and then hit that button to retrieve your angle.

If a right triangle had a = 6 inches, b = 8 inches and c = 10 inches, what would the two missing angles be? The angle opposite of a will be called t and the angle opposite of b will be called s.

s = ____
t = ____

Tip: Every triangle, right or not, is guaranteed to have its angles sum up to 180° if on a flat surface. Therefore, you only need to use trigonometry for one angle, and use arithmetic for the other.

The Monty Hall Paradox: What are the Odds?

2011-07-16T12:00:00.000-07:00

One subject in mathematics always confuses people. No, not Calculus. This is the one some kids call "easy." Probability and Statistics is so difficult for us to understand. Let's look at one of my favorite problems that is famous in the world of game theory: The Monty Hall Paradox.

Suppose you are a contestant on the old TV show Let’s Make a Deal®, hosted by Monty Hall. Monty shows you three doors. Behind two of the doors is trash; behind one of them is a new car. You choose a door, and Monty then opens one of the other doors, revealing trash (he can always do this). You are then given a chance to switch your choice to the other door. What do you do?

The average person will stay because they don't like to be wrong. People also have a strong tendency to go with their gut instinct. Mathematically, the average person also stays because well, it's 50-50 right?

Turns out this is not the case. Switching doors doubles your chance of getting the car, bringing the odds from 1/3 to 2/3. This fact was impossible for me to understand, but I eventually figured out a reason, and understood why humans don't understand statistics.

Let's bring it to the basics, experimental probability. You used these principles to solve June's problem of the week (I hope you did it!!). We will look at all the possible scenarios.

1)You pick the first door and the second door has the car (the second door will always have the car for us). Monty reveals that the third door has trash. Since we said it's better to switch, we switch to door two and get the car.

2) You choose the second door, and Monty reveals that the third one has trash. As a math enthusiast, you know it is best to switch, so you move to door one and see a big pile of junk. Plain, dirty junk.

3) You go to the third door and Monty reveals the first. You then switch to the second door and drive home in a brand new car.

There were three different possibilities, and two got you the car. This got me satisfied. If you have any other reasonings for the people who aren't convinced, please post them! Isn't that cool?!

Divide Almost Any Odd Number into a Number Consisting of all Nines

2011-07-09T12:00:00.000-07:00

Using Discrete Mathematics, we can do so many things. My all-time favorite of them is definitely the amazing proof that states that any odd number that isn't a multiple of five can divide a number consisting of all nines. Sounds hard to believe, but 713 does go into some number consisting of all nines. I'll prove it.

If you think about it, if you divide a number by 713, there are only 713 possible remainders because 713 isn't a remainder. With that in mind, since we are dealing with the numbers consisting of all nines, let's list some out:

9, 99, 999, 9999, ... 9999999...999 (with 714 nines)

Since we have 714 numbers here, two of them must have the same remainder when divided by 713 because there are only 713 possible remainders. Let's take those two numbers, the bigger one is x and the other one is y. Each one of those is equal to 713 times a quotient, two different quotients (say q and p), plus the same remainder (r).

999...999 (with x nines) = 713q + r
999...999 (with y nines) = 713p + r

Technically, we can subtract these two equations from each other. The x nines - the y nines would end up with x - y nines in the front and y zeros at the end because the nines would cancel. According to the Distributive Law, 713q - 713p = 713(q - p) because you can factor a 713 out of both terms. However, the r's at the end cancel each other out, leaving us with no remainder!!! Since q and p are both integers, subtracting them will also be an integer.

999...999 (with x nines) = 713q + r

- 999...999 (with y nines) = 713p + r

999...(x - y nines)...999000...(y zeros)...000 = 713 (q - p)

Since 713 isn't a multiple of 2 or 5, all of the zeros don't matter. Hence, we can cross them all out leaving us with 999...999 (x - y nines) = 713(q - p). Since q - p is an integer, 713 x some integer = a number consisting of all nines. Isn't that cool!!

Fibonacci Day: Addition of Fibonacci Numbers

2011-07-02T12:00:00.001-07:00

As you may have noticed, today is a Fibonacci Day. If you'll notice, it is the second, and two is a Fibonacci Number.

Let's look at a simple pattern within these numbers. What would happen if you add all the Fibonacci numbers up? Infinity, because they go on forever. What if you added Fibonacci numbers, and then stopped at some point. Let's see:

1 = 1
1 + 1 = 2
1 + 1 + 2 = 3
1 + 1 + 2 + 3 = 7
1 + 1 + 2 + 3 + 5 = 12

You might not see a pattern, but there is one. Let's rewrite these sums in a different format.

1 = 1 = (2 - 1)
1 + 1 = 2 = (3 - 1)
1 + 1 + 2 = 4 = (5 - 1)
1 + 1 + 2 + 3 = 7 (8 - 1)
1 + 1 + 2 + 3 + 5 = 12 = (13 - 1)

See it now? Every sum is one less than a Fibonacci number! This pattern will actually go on forever! In order to prove it, we will use something called proof by induction.

If we were to add on the next Fibonacci Number (21), we would also be adding that to the 13 - 1 from before. However, with 13 and 21 being consecutive Fibonacci Numbers, they join together to create the next Fibonacci Number. Since the minus one remains, you always are subtracting one from a Fibonacci Number.

For another proof, let's express each Fibonacci Nber as the difference of the two after it.

(2-1) + (3-2) + (5-3) + (8-5) + (13-8) + (21-13)

You'll see that in the first two expressions, the 2 and -2 cancel out. In the next two, the 3 and -3 cancel. This keeps going until you are left with the greater number in the last expression and the -1 from the first one. That is also a very beautiful proof. Isn't that cool?!

Greatest Common Factor Made Easy: It's Euclid to the Rescue!!!

2011-06-25T12:00:00.001-07:00

In the topic of Number Theory and Discrete Mathematics, you will run onto a topic called "The Greatest Common Factor" (also known as the greatest common divisor). The greatest common factor (abbreviated as GCF) is when you take two numbers and find the highest number that divides into both of them. It basically explains itself.

In school, you learn that to find the GCF of two numbers, make a T-chart with the two numbers on the top, and list out all of their factors below. Then, find the two greatest ones that are the same. However, you don't have to do that according to Euclid, a famous greek mathematician.

Euclid's Theorem is a way to use modular arithmetic to solve the GCF of two numbers. All you do is divide the smaller number into the larger number. Then, take your remainder from that division problem and divide that into the other number. Keep up that process until there is no remainder, and you have your GCF. For instance, what is the GCF of 88 and 33? Well, 88 ÷ 33 = 2 R22, so you now have 22 and 33. 33 ÷ 22 = 1 R11. Since 22 ÷ 11 has no remainder, you have finished, and your GCF is eleven.

Although the how is always cool (it's going to be here), I like to see why these things work. To prove it, I think of how the greeks used to do division without modern day arithmetic: by repeated subtraction. To divide 25 by 4, they would say okay, 25 - 4 = 21, still positive, 21 - 4 = 17, still positive, and keep going until they can't subtract anymore. Since proofs always use variables, we will say we are dividing a by b and the quotient is x, or we subtracted it x times.

Euclid's Theorem is like that, we are looking for the GCF of a and b and when we divide, x is the quotient and (a - bx) is the remainder. Then, you would have b and a - bx. However, we just subtracted a multiple of b. If we added or subtracted more b's, we would end up with the same remainder. So, we can add a multiple of b back, bx, to get what we were looking for: a.

The proof and theorem itself are pretty cool, but it gets better. I'd like you to guess which two numbers between 1 and 100 would take the longest to figure out. A lot of people would immediately say 99 and 98. Well, 99 ÷ 98 = 1 R1, and 1 goes into 98 with no remainder, making 1 the GCF. That would have taken Euclid one step. Then, people might throw 99 and 50 at him, as they are reasonably far apart. Euclid can tackle that one in two steps. 99 ÷ 50 = 1 R49. 50 ÷ 49 = 1 R1. Since 1 goes into 49, one is the GCF of this one as well.

It turns out the answer is 55 and 89. 89 ÷ 55 = 1 R34. 55 ÷ 34 = 1 R21. 34 ÷ 21 = 1 R13. 21 ÷ 13 = 1 R8. 13 ÷ 8 = 1 R5. 8 ÷ 5 = 1 R3. 5 ÷ 3 = 1 R2. 3 ÷ 2 = 1 R1. Since 1 goes into two, we have completed the problem, and found that one is the GCF.

What's so special about 55 and 89 you may ask. They are Fibonacci Numbers, a famous sequence because of its beautiful patterns and attributions to nature. Basically, they start with 1 and 1. Then, Fibonacci added those together to get 2. Then, he added 1 and 2 together to get 3. He continued adding the previous two numbers together going 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, etc. The Fibonacci numbers are so cool that every Saturday that's date is a Fibonacci Number, I will share another pattern within these numbers.

The Problem of the Week Day 5: Week 6/19 - 6/25

2011-06-24T12:00:00.000-07:00

Today is the last portion of our problems! For variables, the easy problem requires them all while the hard one only requires a, b, and q. I would recommend looking over Monday's work for the hard problem if you are doing the hard one because you will require some trigonometry.

Easy Problem: This problem uses some experimental probability. All you do is put the number of what you want over the total amount of objects. For instance, if there was a bag with 1 red jolly rancher and 2 blue jolly ranchers and you wanted to pull out the red one, put 1 red jolly rancher over 3 total jolly ranchers to get 1/3 = 33.3%.

I have a bag with c green marbles and n/z red ones. If I pull out y marble(s), what are the odds that the marble will be green? Express your answer in a percentage.

Hard Problem: To determine the area of a trapezoid, take the two bases and add them together. Then, multiply that sum by the height, and divide that answer by two. This will give you the trapezoid's area.

Hint: To convert inches to centimeters, divide the measurement by 2.54.

1) Convert measurements a and b from Monday to centimeters. On Monday, there were in inches, but we are measuring in centimeters!

Our centimeter conversion for a will be equal to z.
Our centimeter conversion for b will be equal to p.

2) Calculate the height of this trapezoid using sines and trigonometry.

This is the trapezoid you will use to solve the problem.

h = ____

3) Determine the area of this trapezoid. Your answer should be expressed in square centimeters.

If you completed one or both equations, send me your answer. I will tell you if you are correct and list you as one of the people who completed it.

Problem of the Week Day 4: Week of 6/19 - 6/25

2011-06-23T11:07:00.000-07:00

Last week, the easy problem only required the previous day's variables while the hard problem required every variable in that week's history. Today, we are doing it in reverse. For the hard problem, you only need m and n while the hard problem requires c, n and z.

Easy Problem: Today, we will be computing lots of fractions. In order to add fractions, you need a common denominator (the denominator is the bottom number/term). In order to do that, multiply one or both denominator(s) and numerator(s) (top number/term) by a number in order to make the denominators equal. Then, add the numerators and use the common denominator as the denominator of your sum. To multiply fractions, simply multiply the numerators and multiply the denominators.

Plug c, n, and z into this equation and solve for y: [(c + n/z) / 2n] + (1/n + 1/z) c/n = y

y = ____

Hard Problem: When given a negative exponent, all you do is make the exponent positive and find the term's reciprocal (divide the term into 1) For instance, 2^-2 = 1/2^2 = 1/4.

Hint: The graph x^2 + y^2 = r^2 is a circle around the origin (0, 0) with r being the radius of the circle. If you had a less than symbol (<) instead of the equal symbol, shade inside of the circle. For a greater than symbol (>), shade outside of the circle.

1) Solve for g: g = 5(m^-1)(n^-1)

g = ____

2) Graph the following on graph paper: x^2 + y^2 < g

3) Using the formula A = πr^2, find the area of the shaded portion. Since we already used a in our trigonometry, we will make the area = q.

q = ____

The Problem of the Week Day 3: Week 6/19 - 6/25

2011-06-22T12:00:00.001-07:00

Here is Day 3's problem. For the easy problem, you will only need yesterday's answers (keep Monday's for later though). For the hard problem, you will need Monday's and yesterday's answers. Good luck!

Easy Problem: If you have a number right next to an expression in parentheses, that also means multiplication. It's kind of like with a variable!

Plug z in from yesterday's work, and solve for n. [z(4 + z) + 5z]/(3z - 3) = n

Hard Problem: If you have two points on a linear graph, you can figure out it's slope (rate of change) by using the formula y2 - y1 / x2 - x1. So, if you had points (3, 0) and (4, 2), you would do 2 - 0 / 4 - 3 = 2 / 1 = 2. Therefore, 2 is your slope. Then, plug that into y = mx + b as m, and make b your y-intercept (where your line crosses the y axis). If you plug any point from the graph into that equation, it should be correct.

1) If yesterday's x = the x intercept of your graph (x , 0) and y = the y intercept of your graph (0 , y), find the slope (m) of this graph. What is this graph's equation?

m = ____
y = ____x + ____

2) Solve for r in the following equation: 1 + 8r^3 + 4^8 = (2b - a)^3 - 30 • 70 - 4 • 15

3) Plug r in for y in the graph's equation and find x. Since we already used x yesterday, we will call this value n.

n = ____

By the way, if you want to check in and see how you're doing, by all means, contact me at ethan@ethanmath.com. I will get back as soon as possible and tell you if you are correct.

The Problem of the Week Day 2: Week 6/19 - 6/25

2011-06-21T12:00:00.001-07:00

Here is the second part of our first problem of the week! Make sure you remember your answers from yesterday to plug into today's equations!

Easy Problem: To do this problem, you need to do some simple Algebra. Remember to do the opposite operation of what is there. For instance, if you see multiplication, you would do division to both sides to get rid of it. For example, if you had 6 = 2k (a letter with a number to its left without a symbol means multiplication), you would divide both sides of the equation by two to get 3 = k.

Step 1: Plug c into this equation: (2 x c^2) + 1 = ?
Step 2: Make the answer to that equal 17z
In other words, solve (2 x c^2) + 1 = 17z

z = ___

Hard Problem: Since the easy people get one equation to solve, we can handle two! This is a system of linear equations, where you have to solve for two variables, requiring a second equation. In order to solve it, you will multiply through one or both equations by a number that will make two coefficients the same number, but one negative and one positive. Then, you add both equations together, which eliminates a variable and allows you to use simple Algebra to determine the other. After finding that variable, plug its answer into the easiest of the original equations and solve for the other variable. Since the math will be difficult, you may use a calculator for today's work. You will be plugging yesterday's answers into this equation where you see the letter a or b.

-3ax + 156y = 519 - 6b
3bx + 48.2y = 6^3 + .9

x = ____
y = ____

Also, the problem of the week is only going to occur once every month. Yesterday, I said it would be weekly. My mistake. On the first day of the next problem of the week, I will inform you of who got the answer right.

The Problem of the Week Day 1: Week of 6/19 - 6/25

2011-06-20T12:00:00.002-07:00

Since this is the first problem of the week, let me go over how it works. Each week, you have an easy and hard problem to try. The problem is split into five parts, one for each day of the week. If you figure out the answer, contact me and I will tell you if it's right. If it is, I will list you as one of the people who solved this problem. After one month, I will post the answer along with Saturday's post on Cool Math Stuff.

Before you start, don't forget to round all answers you get to the NEAREST TENTH. Tomorrow, you will be solving algebraic equations, and you do not want non-terminating decimals in those!
Easy Problem: When given the two sides of a right triangle, you can use the Pythagorean Theorem to figure out the third. Basically, the longest side is titled c and the others are a and b. Then, you plug the sides into the equation a^2 + b^2 = c^2.

In today's triangle, a = 3 and b = 4. The answer for today's part is c.

Hard Problem: If you only are given one side of your right triangle, but another angle, you can use some simple trigonometry to determine the other two sides. Since the side opposite to your angle divided by c = the sine of that angle, and there is a sine function on some calculators, you can figure it out. All you do is type in your angle, hit the "sine" or "sin" button on your calculator, and multiply it by c to give you another side. Then, use the Pythagorean Theorem to figure out the third.

In our triangle, c = 39.4 in and our angle is 52.5°. Try to figure out a and b (with a being the side you use trigonometry to determine and b being the side you use the Pythagorean Theorem to determine.

Memorizing Times Tables Through Twenty – Without Memorizing Anything!!!

2011-06-18T14:26:00.000-07:00

In school, you probably had to memorize your times tables through ten or twelve. Now, some schools are requiring your times tables memorized through twenty. However, there is a cool method that can make you multiply these numbers instantly in your head, so quick that it seems like you've memorized them! And you didn't!!

Let's take a simple one, 12 x 13. You may know this one to be 156, or already punched it into your calculator. However, we will try to do it anyways. First, you have to add the last digit of the second number to the first number. 12 + 3 = 15. Next, we tack a zero onto this sum to give us 150. Finally, we multiply the two last digits of your number. 2 x 3 = 6. We add this product onto the number we had earlier (150) to give us 156, exactly what we said.

Let's try a bigger one: 18 x 17. First, you add the 18 + 7, the last digit of the second number, to get 25. Then, we tack on a zero to get 250. Finally, we do 8 x 7 = 56. Add that to 250 and you get the answer of 306, which is the correct answer.

With a little bit of practice, you will be able to multiply these numbers faster than a calculator! Isn't that cool?!

1/1	1/2	1/3	1/4	1/5
2/1	2/2	2/3	2/4	2/5
3/1	3/2	3/3	3/4	3/5
4/1	4/2	4/3	4/4	4/5
5/1	5/2	5/3	5/4	5/5

1/1	1/2	1/3	1/4	1/5
2/1	2/2	2/3	2/4	2/5
3/1	3/2	3/3	3/4	3/5
4/1	4/2	4/3	4/4	4/5
5/1	5/2	5/3	5/4	5/5

1/1	1/2	1/3	1/4	1/5
2/1	2/2	2/3	2/4	2/5
3/1	3/2	3/3	3/4	3/5
4/1	4/2	4/3	4/4	4/5
5/1	5/2	5/3	5/4	5/5