Gumball Dilemma

Natasha Oslinger

1.Problem Statement

For this problem, we had to create a set of equations that could be used to calculate how much money one would need to buy a certain number of gumballs of the same color. A woman needed to buy two gumballs of the same same color for her pick daughters, useing my equation, I was able to figure out the most money she would need to spend to get those gumballs.

2.Process Description

I started this problem by reading, then re-reading the information given. Once I fully understood what I needed to do, I made a chart. A labeled one column with number of children that wanted gumballs, the next with how many colors there were, and I left the third column black. This column is for the amount of pennies it would take to buy the necessary number of gumballs. I mmade three of these charts and labeled them 2, 3, and 4. 2, 3, and 4 represent the number of kids, so, chart 2 was to find the number of pennies for gumballs for two children, three for three children, and four for four children. The second column on each chart was for the number of colors of gumballs there were, for each chart, I imputed that there were two, three, four, and five different colors. I played around with the numbers, and figured out that by assuming the worst case scenario, I could figure out the absolute most amount of money that would have to spent. For example, if there are two children and two different colors of gumballs, the most amount of money that would have to be spent is three pennies. I figured this out because if the mom put on one pennie and got one color, then put in another pennie and got another color, the third pennie she puts in whould have to be the same as one of the other two colors. I did this with each number of kids and each number of colors and found the forumlas I needed.

3.Solution

Chart 2

Kids

# of Gumball Colors

Pennies Payed

2

2

3

2

3

4

2

4

5

2

5

6

Chart 3

Kids

# of Gumball Colors (c)

Pennies Payed (p)

3

2

5

3

3

7

3

4

9

3

5

11

Chart 4

Kids

# of Gumball Colors (c)

Pennies Payed (p)

4

2

7

4

3

10

4

4

13

4

5

16

Formula for Chart 2:

c+1=p

I got this formula because I noticed that the number of pennies payed was 3, 4, 5, then 6, a fairly simple consecutive pattern found by adding one to the last number. I was able to create this formula fairly easily. I know this is correct because I plugged in every number for c and always got the correct p. For example, with the number of gumball colors (c) being two, I could add one to two and get three, which is the correct answer. Another example is having c be four, and since four and one is five, I knew my formula was correct.

Formula for Chart 3:

c*2+1=p

I found this formula by noticing another pattern in the number of pennies paid, this time, the answers were 5, 7, 9, then 11. This pattern is created by adding two to the number before it, so, with this in mind, I created the formula. Although this equation is slightly more challenging than the first, it is still correct. For example, when having c be three, I could plug it in and find that 3*2+1 is seven, and when checking my chart, I see that seven is correct.

Formula for Chart 4:

c*3+1=p

I used the same method for this chart as I did for chart 3, only this time, I noticed the pennies paid were being increased by three. Finally, this equation is similar to the one for Chart 3, and like the two before it, it produces the correct answer. For example, if c is five, I know that p is 16. I know this because 5*3+1 is sixteen, which is what I got on my chart.

4.Self Assessment

From this problem, I learned that when there is a pattern, there is a solution. I was able to solve this problem very easily by looking for patterns and seeing how I could make formulas with the information that I already have. For example, when making Chart 4, I wasn’t sure how to go about making the formula. I was able to figure it out very simply by looking at the formula for Chart 3 and changing it slightly, which ended up being a formula that worked perfectly for the problem. I think I deserve a 10 out of 10 on this project. I worked diligently at home and utilized my time in class to finish the problem and the write up. I made sure that all my work was exquisite and check my answers a couple time to make sure I did everything correctly. The Habit of a Mathematician I used most in this project was Staying Organized. To be able to organize data and comprehend given information is a great skill. I was able to finish this project easily and quickly by putting all my information in charts and by keeping on track with what I was doing.

Natasha Oslinger

1.Problem Statement

For this problem, we had to create a set of equations that could be used to calculate how much money one would need to buy a certain number of gumballs of the same color. A woman needed to buy two gumballs of the same same color for her pick daughters, useing my equation, I was able to figure out the most money she would need to spend to get those gumballs.

2.Process Description

I started this problem by reading, then re-reading the information given. Once I fully understood what I needed to do, I made a chart. A labeled one column with number of children that wanted gumballs, the next with how many colors there were, and I left the third column black. This column is for the amount of pennies it would take to buy the necessary number of gumballs. I mmade three of these charts and labeled them 2, 3, and 4. 2, 3, and 4 represent the number of kids, so, chart 2 was to find the number of pennies for gumballs for two children, three for three children, and four for four children. The second column on each chart was for the number of colors of gumballs there were, for each chart, I imputed that there were two, three, four, and five different colors. I played around with the numbers, and figured out that by assuming the worst case scenario, I could figure out the absolute most amount of money that would have to spent. For example, if there are two children and two different colors of gumballs, the most amount of money that would have to be spent is three pennies. I figured this out because if the mom put on one pennie and got one color, then put in another pennie and got another color, the third pennie she puts in whould have to be the same as one of the other two colors. I did this with each number of kids and each number of colors and found the forumlas I needed.

3.Solution

Chart 2

Kids

# of Gumball Colors

Pennies Payed

2

2

3

2

3

4

2

4

5

2

5

6

Chart 3

Kids

# of Gumball Colors (c)

Pennies Payed (p)

3

2

5

3

3

7

3

4

9

3

5

11

Chart 4

Kids

# of Gumball Colors (c)

Pennies Payed (p)

4

2

7

4

3

10

4

4

13

4

5

16

Formula for Chart 2:

c+1=p

I got this formula because I noticed that the number of pennies payed was 3, 4, 5, then 6, a fairly simple consecutive pattern found by adding one to the last number. I was able to create this formula fairly easily. I know this is correct because I plugged in every number for c and always got the correct p. For example, with the number of gumball colors (c) being two, I could add one to two and get three, which is the correct answer. Another example is having c be four, and since four and one is five, I knew my formula was correct.

Formula for Chart 3:

c*2+1=p

I found this formula by noticing another pattern in the number of pennies paid, this time, the answers were 5, 7, 9, then 11. This pattern is created by adding two to the number before it, so, with this in mind, I created the formula. Although this equation is slightly more challenging than the first, it is still correct. For example, when having c be three, I could plug it in and find that 3*2+1 is seven, and when checking my chart, I see that seven is correct.

Formula for Chart 4:

c*3+1=p

I used the same method for this chart as I did for chart 3, only this time, I noticed the pennies paid were being increased by three. Finally, this equation is similar to the one for Chart 3, and like the two before it, it produces the correct answer. For example, if c is five, I know that p is 16. I know this because 5*3+1 is sixteen, which is what I got on my chart.

4.Self Assessment

From this problem, I learned that when there is a pattern, there is a solution. I was able to solve this problem very easily by looking for patterns and seeing how I could make formulas with the information that I already have. For example, when making Chart 4, I wasn’t sure how to go about making the formula. I was able to figure it out very simply by looking at the formula for Chart 3 and changing it slightly, which ended up being a formula that worked perfectly for the problem. I think I deserve a 10 out of 10 on this project. I worked diligently at home and utilized my time in class to finish the problem and the write up. I made sure that all my work was exquisite and check my answers a couple time to make sure I did everything correctly. The Habit of a Mathematician I used most in this project was Staying Organized. To be able to organize data and comprehend given information is a great skill. I was able to finish this project easily and quickly by putting all my information in charts and by keeping on track with what I was doing.