FRACTIONS
The Problem:
Take the fractions ½ and ⅓
Find the fraction between them with the smallest denominator (integers only). Generalize this into a pattern. Extend this pattern for awhile.
Prove it in two parts  First, show that your formula will always choose a number that is between the two original fractions. Second, prove that there are no other solutions between that have smaller denominators.
Challenge for artsy folk  graph your set on a circle. Make it pretty.
Find the fraction between them with the smallest denominator (integers only). Generalize this into a pattern. Extend this pattern for awhile.
Prove it in two parts  First, show that your formula will always choose a number that is between the two original fractions. Second, prove that there are no other solutions between that have smaller denominators.
Challenge for artsy folk  graph your set on a circle. Make it pretty.
"GOOD WILL HUNTING" AND FACTOR SUMS
The Problem(s):
1) The problem from Good Will Hunting was to “draw all homeomorphically irreducible trees of n=10.”
2) Find a number such that the sum of the factors are equal to the number. For example, the factors of 70 are 1, 2, 5, 7, 10, 14, and 35. So, does 1+2+5+7+10+14+35=70? Find as many such numbers as you can.
2) Find a number such that the sum of the factors are equal to the number. For example, the factors of 70 are 1, 2, 5, 7, 10, 14, and 35. So, does 1+2+5+7+10+14+35=70? Find as many such numbers as you can.
TESSELLATIONS
The Problem:
How many different patterns of symmetry exist in tessellations?
Here are some intermediate questions to consider:
 What shapes are possible to create tessellations from?
 How can they be arranged to cover a plane without any blank spaces?
 What types of transformations could the original tile undergo) and still tessellate the plane)?
 How can you categorize these different patterns and make distinctions?
Here are some intermediate questions to consider:
 What shapes are possible to create tessellations from?
 How can they be arranged to cover a plane without any blank spaces?
 What types of transformations could the original tile undergo) and still tessellate the plane)?
 How can you categorize these different patterns and make distinctions?
THE MYSTERY SYMBOL AND 2048
The Problem:Problem 1:
1★1 = 3 a★b = b★a a★(b+1) = a★b + (a+1) + 2*b What is 2017★2018? Problem 2: The Hungarian National 2048 team is hoping to qualify for the Olympics. Unfortunately, 2048 is not a sport. They seem to think they might get it into the Olympics if they can explain the scoring simply. Your goal is to help them develop a formula for the point value of each tile. The input of this formula should be the tile number (1,2,3,4,5...) The output is how many points it is worth. To be clear, the scoring system is already established, you just need to understand it and put a formula to it. 

My Work Explained:
Problem 1:
I started by plugging in different values for a and b into the third equation. I wasn't finding any patterns, so I directed my attention towards what the symbol could mean in the first equation. I wasn't finding any patterns. I spoke with peers and decided to try by using the third equation and plugging in what I already knew. I found that  1★2 = 7 , 1★3 = 13 , 1★4 = 21 , 1★5 = 31 and so on. I found that the pattern was that it was adding only even numbers from the last solution. For example, 7+6 = 13, 13+8 = 21, 21+10 = 31 and so on. I decided to graph this and found a nice parabola with the equation x^2+x+1. Using this equation, I found that 1★2018 = 4,074,343. That's where I stopped on Friday.
Problem 2:
I did not work on this problem.
I started by plugging in different values for a and b into the third equation. I wasn't finding any patterns, so I directed my attention towards what the symbol could mean in the first equation. I wasn't finding any patterns. I spoke with peers and decided to try by using the third equation and plugging in what I already knew. I found that  1★2 = 7 , 1★3 = 13 , 1★4 = 21 , 1★5 = 31 and so on. I found that the pattern was that it was adding only even numbers from the last solution. For example, 7+6 = 13, 13+8 = 21, 21+10 = 31 and so on. I decided to graph this and found a nice parabola with the equation x^2+x+1. Using this equation, I found that 1★2018 = 4,074,343. That's where I stopped on Friday.
Problem 2:
I did not work on this problem.
TRIGSTAR PROBLEMS
The Problem:A collection of problems that require trigonometry to solve them.
My Work Explained:I solved the problems using the Law of Cosines, Law of Sines, formula for arc length, Pythagorean theorem, and area formulas.


GAMES OF 100
The Problem:LOADING THE AIRPLANE  100 people are boarding an airplane, each with an assigned seat. However, the first person on board loses their ticket and decides to just sit wherever they'd like. Each successive person to board the flight will go to their assigned seat. If it is open, they'll sit down. If it is occupied, they'll take another seat at random. No one has the gall to confront a stranger and make them move seats. You are the last person to board the plane. What is the probability that you will get your own seat? What is the probability that only two people are in the wrong seat?
LONG LINE OF CARS  100 cars are lined up in a random order to drive from Durango to Cortez. Each car drives at a constant speed, but everyone's speed is different. With road construction, no one is able to pass for the whole trip. Therefore, if you get placed behind a car that is slower than you, you have to slow down and match their speed. In how many groups would you expect to arrive? 

LOGICAL CONFERENCE  100 people with a gift for perfect, infallible logic attend a conference. They each have a hat placed on their head so they can see everyone else's hat, but not their own. They are then given the following instructions from the game master: "Without communicating with each other in any way, your goal is to determine if your hat is blue or orange. In each round, you will look around and observe your fellow logicians. At the end of each round, I will turn the lights off. If you know the color of your own hat, you must leave the room. If not, I will turn the lights back on for the next round. At least one of you in wearing a blue hat." What he kept to himself was that everyone was wearing a blue hat. So each person looks around and sees 99 other people wearing a blue hat, but wonders silently about the color of their own hat. Who leaves the room, and when?
My Work Explained:
#1  I started out by finding each person’s probability that their correct seat will be taken. I found that the first person to board (without a ticket) has a 0% chance that their seat is taken. The second person has a 1% chance that their seat is taken, because person #1 could sit in person #2’s seat. Person #3 has a 2% chance, person #4 has a 3% chance, and so on. We also found that the first person has a 1% chance of sitting in the seat that was on their ticket. In that case, everything would be fine and go as planned. I realized that finding the probability all 100 people was going to be too overwhelming and decided to break it down, so we started with 5 people. I created a probability table. After that, I went to 10 people. We created another probability table. After I created these tables, we looked back at the main questions to answer the first one. If you consider 5 people, the last person has a ⅕ chance of getting their seat. If you consider 10 people, the last person has a 1/10 chance of getting their seat. We assumed that with 100 people, the last person would have a 1/100 chance of getting their seat. I then attempted to answer the second question. What is the probability that only two people are in the wrong seat? I decided to take a similar approach by starting with smaller groups. I knew that with 2 people, there are only two possible scenarios: They sit in their correct seats, or they sit in the other’s seat. Therefore, the probability that 2 people sit in the wrong seat is 50%.
With 3 people, I knew this: (Only a possible scenario.)
Out of these 5 possible scenarios, only 2 of them show only two people who were sitting in the wrong seat. So the probability that only two people are in the wrong seats for a group of 3 people is: ⅖.
That’s where I stopped for the day. I was about to find all the possible scenarios for 5 people, and the probability for 4 people.
#2  We figured these things out:
With 3 people, I knew this: (Only a possible scenario.)
 The probability that they all sit in their correct seats is ⅓.
 The probability that person #1 sits in person #2’s seat is ⅓.
 The probability that person #2 sits in person #1’s seat is ½.
 The probability that person #3 sits in person #3’s seat is 1/1.\
 This is only possible is 2 people switch seats. For example, #3 sits in #5’s seat and #5 sits in #3’s seat.
 Out of the 2 people that switch seats, one of them must be person #1.
 The other person must sit in person #1’s seat. Therefore, person #1 must sit in the other person’s seat, which creates 99 possible seats for that person to choose and 1/99 chance that the person chooses person #1’s seat.
Out of these 5 possible scenarios, only 2 of them show only two people who were sitting in the wrong seat. So the probability that only two people are in the wrong seats for a group of 3 people is: ⅖.
That’s where I stopped for the day. I was about to find all the possible scenarios for 5 people, and the probability for 4 people.
#2  We figured these things out:
 The perfect scenario would be that the fastest car is first, the second fastest car is second, and so on. This scenario would create 100 different groups.
 If the slowest was in front, it would create 1 big group.
 If the slowest was in the middle, it would create 251 groups, depending on the other car’s speeds.
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