Maths Write-Up Problem Statement: (Restate the problem using words, pictures, and/or a diagram)
(x) amount of children have encountered a gumball machine. That gumball machine has (y) different colors of gumballs. Those children all want the same colored gumball. Each gumball costs 1 cent, so how much will it costs to get (x) children the same colored gumball.
Process Description: (How did you try to solve the problem? You may also consider how others in your group tried)
To be honest, my first thought on this problem was an unusual mistake. I thought that Ms. Hernandez would be putting in an infinite amount of pennies because she would get the same color over and over again. I knew it was wrong though because the most she could spend was 3 cents. I don’t know why this came to mind but thankfully I discarded that idea quickly. Then I thought about why the most she would have to spend was 3 cents. I started thinking about all of the possible outcomes and how much Ms. Hernandez would have to spend based on each situation. Since the 2 children don’t care as to what color gumball they get, I thought of 4 possible outcomes. The first one is if Ms. Hernandez gets 2 red colored gumballs in a row, only spending 2 cents. The second one is if Ms. Hernandez gets 2 white colored gumballs in a row, only spending 2 cents. The third one is if Ms. Hernandez gets 1 red colored gumball, 1 white colored gumball, and then 1 more red gumball, only spending 3 cents. The fourth and final outcome was if Ms. Hernandez gets 1 red colored gumball, 1 white colored gumball, and then 1 more white gumball, only spending 3 cents. I then visualized all of these outcomes and carefully examined each one. By then, I knew why Ms. Hernandez would have to spend 3 cents at most.
Later on I realized that I had not thought of 2 outcomes, so in total there were 6 outcomes. These would have not changed much and it would of not affected the answer in the first place but, they were important to look back on. The fifth outcome was if Ms. Hernandez got 1 white gumball, 1 red gumball, and then 1 more red gumball, only spending 3 cents. The sixth outcome was if Ms. Hernandez got 1 white colored gumball, 1 red colored gumball, and 1 more white colored gumball.
I used this possible outcome method for questions 2 and 3. I used them on these questions because questions 2 and 3 used the same components as the first question.
Number of children (x) Number of gumball colors (y) Max number of pennies (p)
2 2 3
2 3 4
3 3 7
4 4 13
x * y + 1 - y = p
Equation: x * y + 1 - y = p
Extensions: Invent some extensions or variations to the problem; that is, write down some related problems.They can be easier, harder, or about the same level of difficulty as the original problem. (You don't have to solve these additional problems.)
One problem I thought of was if there were 4 children that wanted the same color gumball. These children also didn’t care as to what color gumball they got, as long as they all got the same color. There were 4 different colored gumballs in the machine. Those colors were red, white, blue, and green. Each gumball cost 5 cents and I wanted to find out how much you would have spend in order to meet the children’s needs. After thinking of all of the possible outcomes and discussing with my group, I came to the conclusion that you would have to buy 13 gumballs at most and that would cost 65 cents.
Solution: (The end result, wrong or right, it doesn’t matter! Include one or many solutions as long as they make sense to you!)
In the end, I FOUND a formula that solves problems 1-3 and even my own problem. I tried multiple equations like x*y + 5 divided by 3 = p, x*y + 8 - 6 = p, and others. Each one of these equation solved 1 or 2 of the questions but not the other. After countless tries, I came up with the equation, x * y + 1 - y = p which worked for each equation.
For example,
Problem 1 is solved like this: 2 * 2 + 1 - 2 = 3 2 * 2 = 4 + 1 = 5 - 2 = 3
Problem 2 is solved like this: 2 * 3 + 1 - 3 = 4 2 * 3 = 6 + 1 = 7 - 3 = 4
Problem 3 is solved like this: 3 * 3 + 1 - 3 = 7 3 * 3 = 9 + 1 = 10 - 3 = 7
My own problem is solved like this: 4 * 4 + 1 - 4 = 13 4 * 4 = 16 + 1 = 17 - 4 = 13
Self-Assessment: Reflect on one Habit of a Mathematician you used when solving this problem
One Habit of Mathematician I used while solving this problem was Describe and Articulate. When I thought of a particular outcome, I visualized it in my mind and compared it to my other outcomes. I did this to every outcome I could think of for this particular problem. This really helped me a lot because I am a visual learner. I will definitely use the Habit of Mathematician in the future with other problems because it connects to my strengths and gives me a better understanding of the problem.
Problem of the Week #2 Matthew Evans 2016 Maths Write-Up Problem Statement: (Restate the problem using words, pictures, and/or a diagram)
Here we have a game that has (x) sticks. You can cross out up to (y) sticks. This game can only be played by 2 people. The winner of the game is decided by the person that crosses out the last stick to where every stick has been crossed out. Think of strategies for a certain situation you’re in while playing the game. What strategy or strategies have you come up with and how does it vary depending on different situations?
Process Description: (How did you try to solve the problem? You may also consider how others in your group tried)
Well at first my group and I decided to experiment by playing with 4 players instead of 2. We also added 10 more sticks to play with 20 instead of 10. After we played once and got used to the mechanics of the game, we played in pairs and everyone played with each other (I won every game that I played! :D). The first person I played against was Josh. After I had beat him, I found 2 strategies that would help you as the first player. As player 1, you would only cross out 1 stick. Then if they cross out 2 or 3, you have the advantage and can win the game without any trouble, as long as you cross out the amount of sticks needed for victory as shown below.
1st Strategy as Player 1: You cross out 1 stick (9 sticks left), They cross out 2 sticks (7 sticks left), You cross out 3 sticks (4 sticks left), They cross out 1, 2, or 3 sticks (3, 2, or 1 sticks left), You cross out 1, 2, or 3 sticks (0 sticks left).
2nd Strategy as Player 1: You cross out 1 stick (9 sticks left), They cross out 3 sticks (6 sticks left), You cross out 2 sticks (4 sticks left), They cross out 1, 2, or 3 sticks (3, 2, or 1 sticks left), You cross out 1, 2, or 3 sticks (0 sticks left).
The next player I played against was Fernando, whom I won against. I could not find a strategy or variation out of our game and in fact, the only reason I won was because Fernando made the wrong move towards the end of the game, which had cost him it. Lastly I played against Mariela, whom I also won against. This game had not decided a winner until one of the very last moves. By that move, I had seen how many sticks were left and crossed out the correct amount to secure my victory. I could find any strategies or variations out of this game as well. After that, I couldn’t find another strategy that had exposed itself within our games.
Extensions: Invent some extensions or variations to the problem; that is, write down some related problems.They can be easier, harder, or about the same level of difficulty as the original problem. (You don't have to solve these additional problems.)
I’m not really sure how or why I thought of this problem. There are 15 balloons that are circling each other in a very small amount of space. You play with another person and use a bow and arrow each round. You only have one shot before it is the next person’s turn. You’re goal is to pop the last balloon and the order of encountering that balloon can be changed by popping more than one balloon with your one arrow each round. Find strategies within who might be the player to hit the last balloon. Is this affected by luck or skill?
Solution: (The end result, wrong or right, it doesn’t matter! Include one or many solutions as long as they make sense to you!)
In the end, I only found 2 strategies that were variations of the same strategy. Also, if I had to choose between being Player 1 or Player 2, I would choose to be Player 1 because of the strategy that I found. It gives me reason for wanting to go first and also because after you and the other person go once, you have great control over the game and how it might result in. Player 2 does get the advantage of not going first and gets to see how their opponent might play out the game and use their strategies. That still isn’t enough to make me want to be Player 2 though. If we were to change the amount of sticks to 20 or something higher than 10, it would actually give the second player a chance to change the course of the game. This is why the strategy I found mainly works for a few situations while only having 10 sticks in total.
Self-Assessment: Reflect on two Habits of a Mathematician you used when solving this problem
One habit of Mathematicians I used was Generalize. I used this while coming up with the different ways of using the strategy I found while playing Nim against Josh. I grew in using all of the different variations and explaining them in small clear steps. Another habit of Mathematicians I used was Stay Organized. I used this while separating my work and my clear steps within this document. I grew in explaining everything while I wrote down the different variations of my strategy.
Problem Of the Week #3 The Broken Eggs
Your Task Your task is to answer the insurance agent’s question. In other words, What can the farmer figure out from this information about how many eggs she had? Is there more than one possibility?
Write Up
Problem statement - re-write the question in your own words
Process Description - describe and show how you and your group attempted to solve the problem
Extensions - invent some extensions of variations of the problem
Solution - describe your final solution to the problem
Reflection - reflect on how you used two habits of mathematicians to solve the problem.
Write Up Maths Write-Up Problem Statement: (Restate the problem using words, pictures, and/or a diagram)
A farmer is bringing her eggs to the market in a cart. On the way, she hits a pothole and all of her eggs drop out of the cart. Every egg broke and her trip is rendered useless so she goes to the insurance company and tells them what happened. They ask her how many eggs she had while going to the market. She doesn’t remember how many she had but she knows that when she put the eggs in groups of 2, 3, 4, 5, or 6, there was always 1 egg left over. When she put the eggs in groups of 7, there were no eggs leftover. How many eggs did the farmer have in total?
Process Description: (How did you try to solve the problem? You may also consider how others in your group tried)
At first, I came to the conclusion of using the multiples of 7 to calculate which number corresponds to the number of eggs she had. I saw that it would take forever if I did it this way so I narrowed down my choices to odd numbers and numbers that relate with multiples of 5. For example I wanted a number that’s 1 greater than a 5 multiple and an odd number. After searching through all of the numbers I narrowed my search to, I found that 301 had matched the number that needed to fit the eggs in groups the way the farmer had described it. She had 301 eggs because when you put the eggs in groups of 2, 3, 4, 5, or 6, there is always one egg leftover. When you put 301 eggs in groups 7, there are a perfect amount of eggs in each group.
301 Eggs in Groups of 2: 2 x 150 groups = 300 eggs + 1 leftover egg = 301 eggs
301 Eggs in Groups of 3: 3 x 100 groups = 300 eggs + 1 leftover egg = 301 eggs
301 Eggs in Groups of 4: 4 x 75 groups = 300 eggs + 1 leftover egg = 301 eggs
301 Eggs in Groups of 5: 5 x 60 groups = 300 eggs + 1 leftover egg = 301 eggs
301 Eggs in Groups of 6: 6 x 50 groups = 300 eggs + 1 leftover egg = 301 eggs
301 Eggs in Groups of 7: 7 x 43 groups = 301 eggs
Extensions: Invent some extensions or variations to the problem; that is, write down some related problems.They can be easier, harder, or about the same level of difficulty as the original problem. (You don't have to solve these additional problems.)
One extension I have to this problem is that there are 179,047 ants in a huge container together. The person counting these ants wants to put these ants together in groups of equal amounts of ants. How many ants are in a single group and how many groups of ants are there overall?
Solution: (The end result, wrong or right, it doesn’t matter! Include one or many solutions as long as they make sense to you!)
In the end, I found the solution of the farmer having 301 eggs in total before they all broke. This number perfectly matched the description the farmer gave us while she tried to remember how many eggs she had. It’s puts 301 eggs into groups of 2, 3, 4, 5, and 6 with one one egg leftover for each of these different groups. When we put 301 eggs into groups of 7, the groups are complete with not a single egg leftover. With this much evidence, I know that 301 eggs is the answer to the farmer’s problem.
Self-Assessment: Reflect on two Habits of a Mathematician you used when solving this problem
One habit of mathematician I used while working on this problem on the week is Start Small. I started with looking at the multiples of 7 to find my answer but ended up coming up with restrictions for numbers I looked into. You definitely need to be detailed while looking for your answer because it shows you fast and efficient ways to find your answer. Another habit of mathematician I used while solving this problem was Conjecture and Test. I thought of rules a number had to have so I could look into it and see if that was how many eggs the farmer had. This was another great way of finding my answer in a more efficient way.
Problem of the Week #4 - 1-2-3-4 problem
You’ve seen that you can change the meaning of an arithmetic expression by inserting or removing parentheses. Of course, another way to change the meaning of an expression is to rearrange its terms.
This problem is about using the digits 1, 2, 3, and 4, in any order you choose, to create arithmetic expressions with different numerical values according to the rules for order of operations.
For this problem, a 1-2-3-4 expression is any expression written using each of these digits exactly once, according to the following rules:
You may use any of the four basic arithmetic operations—addition, subtraction, multiplication, and division (according to the order-of-operations rules). For example, 2 + 1 • 3 – 4 is a 1-2-3-4 expression for the number 1 (since 2 + 1 • 3 – 4 = 1).
You may use exponents. For example, 23 – 4 – 1 is a 1-2-3-4 expression for the number 3.
You may use radicals. For example, is equal to 3, so is a 1-2-3-4 expression for the number 6.
You may use factorials. For example, 4! means 4 • 3 • 2 • 1, so 3 + 4! + 1 – 2 is a 1-2-3-4 expression for the number 26.
You may juxtapose two or more digits (that is, put them next to each other) to form a number such as 12. For example, 43 – 12 is a 1-2-3-4 expression for the number 31.
You may use parentheses and brackets to change the meaning of an expression. For example, according to the rules for order of operations, 1 + 4 • 32 is a 1-2-3-4 expression for the number 37. You can add parentheses and brackets to get [(1 + 4) • 3]2, which is a 1-2-3-4 expression for the number 225.
Your Task: Your task in this problem is to create as many 1-2-3-4 expressions as you can for each of the numbers from 1 to 25. Remember: In every case, the expression must use each of the digits 1, 2, 3, and 4 exactly once.
Write Up
Problem statement - re-write the question in your own words
Process Description - describe and show how you and your group attempted to solve the problem
Extensions - invent some extensions of variations of the problem
Solution - describe your final solution to the problem
Reflection - reflect on how you used two habits of mathematicians to solve the problem.
Problem of the Week #4 Matthew Evans
Maths Write-Up Problem Statement: (Restate the problem using words, pictures, and/or a diagram)
As you’ve seen, you can change as to what an arithmetic pattern means by removing or inserting parentheses. There is another ways to change the meaning though, with that being changing the terms in the equation. In this problem you can use the numbers 1, 2, 3, and 4 to create an equation. There are multiple terms use can use like addition, subtraction, multiplication, division, exponents, radicals, factorials, and parentheses. You may also put numbers together to make a different number (12 + 3 - 4). You may only use these numbers once in the equation and for each equation. Try to find multiple equations that use the numbers 1, 2, 3, and 4. Make sure that you find one equation that answers the numbers 1-25.
Process Description: (How did you try to solve the problem? You may also consider how others in your group tried)
At first, me and my group just came up with a few equations and showed each other to make sure it was right. As soon as we had started individually thinking, I knew that there were so many equations to be found because just changing one number of term changes the entire equation.
Then I looked at the board and saw that Dr. Cate had written the numbers 1-25 on the board. Dr. Cate told us that we had to make a 1-2-3-4 equation that equaled one of these numbers. Only one person could write their equation next to a number. As soon as she finished saying the instructions, people in my class started writing beside the numbers. I really wanted to put something so I found an equation on my paper that equaled a number no one had put up yet which was the number 15. The equation I put up was 4 ÷ 2 + 13 = 15.
After the fiasco of making everything right was finished, we had found 25 solutions to Problem of the Week. Below is a picture with the 25 problems the class found for each number:
Some equations I found were:
1 + 2 + 3 + 4 = 10
4 x 3 ÷ 2 - 1 = 5 4 x 3 ÷ 2 + 1 = 7
3 x 4 x 1 ÷ 2 = 6
(4 x 1) x (3 - 2) = 4
2 x (4 x 3) + 1 = 25
(4 - 3) x 2 + 1 = 3
4 x 3 x 2 - 1 = 23
(4 x 2) + (3 + 1) = 12
4 x 3 x 2 x 1 = 24
4 x 1 - 3 + 2 = 3
4 - 1 - 3 + 2 = 2
1 + 2 - 3 + 4 = 4
Extensions: Invent some extensions or variations to the problem; that is, write down some related problems.They can be easier, harder, or about the same level of difficulty as the original problem. (You don't have to solve these additional problems.)
An extension to the problem I made was to use only words to describe the equation. You can choose whatever numbers and terms but you have to describe it in complete sentences. We worked on doing this while working on functions and describing what our functions meant.
Solution: (The end result, wrong or right, it doesn’t matter! Include one or many solutions as long as they make sense to you!)
Overall, I (with help from the class) found 38 equations to answer this problem. All of these equations follow the requirements and restrictions on the problem and I know that there are plenty more to be found. Now I know though that there are plenty of ways to find an equation for a number yet alone an equation for every number. I also would like to incorporate my extension to the problem somehow to make this problem a lot more understandable to the problem solver.
Self-Assessment: Reflect on two Habits of a Mathematician you used when solving this problem
A Habit of Mathematician I used while working on this problem was Conjecture and Test. I tested different equations and either found them to be within the requirements or not and that’s when I made adjustments to the equation so it would fit the requirements. It’s more efficient to keep on going and revise your problem and not just start over. Another Habit of Mathematician I used was Start Small. I used this while starting off with equations I knew and branched other equations off of it. This is a great way to explore new areas by deeply researching and looking through one area.