9/12/10
IMP
POW Linear Nim
In this POW, we had to play a game called Linear Nim. In this game, we drew 10 lines on a paper, and we had to take turns crossing out 1, 2, or 3 of the marks. The person that crossed out the last mark was the winner. The first task of this POW was to find a winning strategy for this game. After we found this out, we were supposed to make variations to the game, for instance starting with more or less marks, or allowing a player to cross out more or less marks. We were supposed to consider a variety of examples and look for generalizations in strategies. When I first played the game, I didn’t really have any initial strategy in mind. I just picked numbers at random to choose until getting down to the last few. I thought this game was very simple and easy to understand. One key insight I had was to cross out three marks to get rid of most of the marks. When it came down to the final three or four marks, I had to look carefully at them to see what numbers I could choose before the next person’s turn, so after their turn there would be one, two, or three marks left. For instance, if there were 5 marks left, I could cross out one, and then any number the next person picked would leave me with a number of marks I could cross out with one, two, or three crosses. This is when I realized the strategy of this game. The strategy I developed for the original game was that to win, you have to cross out a number of marks so that there are four marks left before the next persons turn. If your opponent picks three, there will be one mark left, and you can cross it out and win. If your opponent crosses out two marks, there will be two marks left, and you can cross those out and win. If your opponent crosses out only one mark, there will be three marks left, and you can cross those out and win. If you get down to four marks before the next person’s turn, there is no possible way you can lose. At least without any variations. My first variation was...

...Taylor Gray
1/31/11
POW 6: linearnim
For this game of LinearNim you draw 10 line marks on a piece of paper and two players take turns crossing off only 1, 2, or 3 marks per turn. The person who crosses off the last mark is the winner. Firstly what I did was play a few games with my Mom and what I realized right away was that if you stopped just before the last four dashes in the game then you would always win. Since you aren’t always guaranteed of being the one who can put these dashes on the sixth line then I had to think of a strategy. So I thought of a way to make it so I would be able to get the sixth dash as long as I started first. I found that starting with two lines crossed off would leave my mom with only one, two or three dashes that would give me the opportunity to get the sixth line no matter which number of lines she crossed off. I used this many times on her and it seemed to be undefeatable. When she went first and didn’t start with three I would try to get the six, which would make me win. Such as if she started with three then I would cross off three and I would get the six forcing her to lose.
For my variation of different games I did 11 lines and 23 lines. I chose eleven because it is right after ten. I chose 23 because it is my sister’s lucky number. I did both of these a few times to find a strategy that would fit any type of variation of linear...

...Problem statement. There’s this game called linearnim where 2 players who have 10 marks and so they have to figure out a strategy. Then who ever crosses out the last mark wins. You can also play it with 15 marks. But you have to figure what to do while playing this game and try to find patterns or strategies to win.
Process. So what I did to attempt the problem is that I played the game a few times with my partner with the 10 marks and 15. So we can find some patterns and strategies that we can discuss. Then some strategies that we came up with were that we had to think ahead of our partner and see what next move will they make or see the amount of marks are left over.
Solution. So while playing the game my partner had won every game. I could see she had a great strategy and we had took turns like if she went first then I would go first the next round. Then I saw that if we played 10 marks I had crossed out 2 she would cross out 3 then 5 were left over so I couldn’t cross out 4 or 5 only 1,2 or 3 . So I cross out 2 and she would cross out the rest. But when we played 15 I could see that if there were 5 left and I have crossed out 1 she could cross out the other 4.
Extension. Mary and Louie are playing the same game but there are 20 marks so they can only mark out even numbers and the last person to cross out the last even number wins.
Self Assessment. My strongest points were that you could think ahead . Also seeing what marks were...

...statements. Together it says that no birds less than 9 feet tall are in the aviary. Then when you combine that statement with the fourth statement you deduce that the birds in the aviary are ostriches. And when you conclude the third statement you get a final statement that says the birds in the aviary are ostriches and they do not eat mince pies.
Part Two
Valid Conclusion: People with hats are untrustworthy
Untrustworthy people are dangerous.
Invalid Conclusion: Some Golden Retrievers are gold
These dogs are gold
3. This POW has to do with mathematics because it is all about logic. In math without logic you would less successful then if you do have logic. If something does not seem right to you or is confusing you are using logic to realize that something is wrong and will use it to solve it. Without logic you wouldn’t know when something is right or wrong.
4. This POW was fairly simple. But some of the problems required you to think more than others. But if you think it out in steps you will be able to solve them easily. I believe that I did well and thought out each problem as well as I can....

...“A Sticky Gum Problem” POW 4
Problem statement:
The next scenario is very similar. In this one, Ms. Hernandez passed a different gumball machine the next day with three different colors Once again her twins each want a gumball of the same color, and each gumball is still one cent. What is the most amount of money that Ms. Hernandez would have to spend in order to get each of her daughters the same color gumball?
In the last scenario, Mr. Hodges and his triplets pass the same gumball machine that Ms. Hernandez and her twins passed in scenario two. This time, each of Mr. Hodges children wants the same color gumball out of the three-color gumball machine. What is the most amount of money that Mr. Hodges would have to spend on his triplets in order to get them each the same color gumballs?
Process:
In the process of solving the first question I drew up different color gumballs.
These are both the different colored gumballs. If this was the outcome after spending two pennies on two gumballs, then the next gumball would have to be one of the previous color gumballs that already came out of the machine.
For the next scenario it was a little trickier that the first problem. Since there were still only two children involved and there were three colored gumballs it wasn’t too hard. Once again, I drew up the three different colored gumballs in the gumball machine. The gumballs were red, white, and...

...Mega POW
A very wealthy king has 8 bags of gold, which he trusts to some of his caretakers. All the bags have equal weight and contain the same amount of gold, all the gold in the kingdom. Although, the king heard a story that a woman received a gold coin. The king knew it had to be his gold so he wanted to find the lightest bag in the 3 weighing, but the mathematician thought it could be done in less, so I need to find out the least amount of weighing it takes to find the lightest bag. Also, the king used a pan balance for all of his weighing.
I started by weighing 4 bags on each side of the scale to see which side was lighter. Then from those results I thought to weigh the 4 bags that were on the lighter side by 2 and 2. After this you would find one side weighing less than another. Then you would take those results and weigh the 2 remaining bags and the lightest bag would be the bag that was taken from. However, the mathematician said it could be done in less than three steps. So throwing the answer I had just gotten to the side, I started new. This time I started with 3 bags on each side knowing that if two sides were equal than the bag with the missing gold would be one of the bags not weighed the first time. Then you would have to weigh the two remaining bags and whichever one was lighter than the other would be the bag with less gold. But, if the 3 bags from the beginning weighed different then you would weigh 2 bags of the 3 and if they are equal...

...Sage Wentzell-Brehme
Problem Statement
In this problem I am looking at strategies for a game called What’s on Back?
This game has 3 cards. One card has an X on both sides, one card has an O on both sides, and one has an O on one side and a X on the other. On your turn you choose one of the cards at random from a bag. You hold it up with one side facing you. You are not allowed to look at the other side or the other cards. You have to predict whether it is an O or an X on the other side of the card.
My task is to look at a bunch of different strategies for playing and find the probability of success for each one. For each strategy I look at I have to find the probability using both an experimental and theoretical model. My real goal is to find the strategy with the highest probability of success or the one most likely to help me win.
Strategy # 1
a. Always choose the same thing the card says. So if it is an O choose O, if it is an X choose X.
b. 30 trials
1. yes 6.no 11.no 16.yes 21.yes 26. yes
2. yes 7.yes 12.yes 17.yes 22.no 27. yes
3. yes 8.no 13.yes 18.no 23.yes 28. yes
4. no 9.no 14.yes 19.yes 24.no 29. yes
5.no 10.yes 15.yes 20.no 25.no 30. Yes
P (right) – 18/30 or 6/10 or .6
Success rate = .6
c. (XX)-works-1/6
(XX)-works-1/6
(OO)-works-1/6
(OO)-works-1/6
(OX)-doesn’t work
(XO)-doesn’t work total successful – 4/6 or...

...Problem Statement: A football league scores points for field goals as five points and touchdowns as three points. They want to know what the highest amount of points that is impossible to score.
Process: I thought that maybe there was patterns in certain combinations.
Field goals = 5 pts. Then I listed the multiples of 5 to 25.
5,10,15,20,25
Touchdowns = 3 pts. I did the same as above.
3,6,9,12,15,18,21,24,27
I looked at the patterns and knew that if the team only scored field goals they would go up by 5 pts. each score. If the team only scored touchdowns they would go up by 3 pts. each score. I decided that I was going to keep looking for patterns in the numbers for different combanations ex. one field goal the rest touchdowns. To keep track of the patterns i was going to make a chart 1-100 of all the patterns and numbers.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100
Touchdown rest Field Goals = 3,8,13,18,23,28,33
I noticed that one the digits got into the teens there was a pattern of hitting all the numbers ending with 3 and 8.
Field Goal rest Touchdowns = 5,8,11,14,17,20,23,26,29
This pattern is a little bit more tricky, but i noticed that its just like the threes it goes up three every number...

...POW 8
Problem Statement-
For this POW, our task was to find the best formula for finding the area of any polygon that is formed on a geoboard. In order to do this, there are two formulas given to help you. One tells how to get the area of a polygon based on the number of pegs on the boundary. This works as an In-Out table, where In is the amount of pegs on the boundary, and Out is the area. The other formula tells how to get the area by having a polygon with exactly four pegs on the boundary. All you do is tell her how many pegs are on the inside. Using her formula, she can give you the area immediately. Using these two formulas for reference, we have to find a formula that anyone can use to find the area of any polygon by knowing the number of pegs on the boundary and interior.
Process-
The first two equations by Freddie and Sally were a preparation for the final, building up the information in order to find the "superformula".
For Freddie's equation, I drew an In-Out table, with the number of pegs on the boundary for the In/x value, and the area of the polygon for the Out/y value. With this table set up, I made some different polygons on geo-paper (attached), and plugged them into the table. I found the area for all of them by looking at the figure. However, just looking at the figure was not very accurate, so I looked for a pattern between the in and the out, and quickly found a formula that made sense, and I worked...

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