More Patterns at CTY: All in one triangle!!

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

First off: Pascal's Triangle

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

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

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

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

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

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

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

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

Bonus: Pascal Magic

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

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

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

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

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.

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 = ___

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.