Saturday, July 30, 2011

More Patterns at CTY: All in one triangle!!

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

First off: Pascal's Triangle

1 1
1 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?

Saturday, July 23, 2011

Patterns and Puzzles at CTY

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

Friday, July 22, 2011

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

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.

Thursday, July 21, 2011

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

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 = ___

Wednesday, July 20, 2011

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

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.

Tuesday, July 19, 2011

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

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!!

Monday, July 18, 2011

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

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.

Saturday, July 16, 2011

The Monty Hall Paradox: What are the Odds?

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?!

Saturday, July 9, 2011

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

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!!

Saturday, July 2, 2011

Fibonacci Day: Addition of Fibonacci Numbers

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 Number 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?!