Amber Truhanovitch                                                                                                                                      Dec. 12, 2007

POW 3                                                                                                                                                                  Group-C

Sticky Gum

http://tbn0.google.com/images?q=tbn:lTa_4aIv3WRUnM:http://www.sxc.hu/pic/m/i/ic/icekitty37/346968_gumballs.jpg

Problem Statement:

                In Sticky Gum Problem we must find a function that will produce the total cost a parent will have to pay if their children want the same color gum ball from a gum call machine. The function must have an input of the number of children and the number of different colors in the gum ball machine. The output must be the total cost in cents to a parent.

Process:

                The first place to start is with the problems given. We need to find how much a parent will pay if they have two children and the gum ball machine has only two different colors and then a machine with three different colors. Then we must find out how much they will pay if they have three children and a gum ball machine with three different colors.

                So first we know that there are two children and there are two colors in the gum ball machine. Each gum ball costs one penny.

                So the first penny is used and one color gum ball is returned.

                The second penny is used and if the parent has bad luck the ball will return a different color.

  1 ˘      Red

  2 ˘      Blue

  3 ˘     Red or Blue

= a match and 3 ˘ spent

 
                So with two children and each penny returns a different color, the third penny with match one of the two colors so there will be a match and children will have the same color gum balls. The parent will have ended up using three cents. To show this I drew a diagram/ chart below.

 

 

  1 ˘      Red

  2 ˘      Blue

  3 ˘     white

  4 ˘   red, white or blue

= a match and 4 ˘ spent

 
 

 


I did this same process for the two children and three gum balls. The parent pays 4 ˘

 

 

 

  1 ˘      Red

  2 ˘      Blue

  3 ˘     white

  4 ˘     red

  5 ˘     blue

  6 ˘    white

Now only two matches and the 7th cent is one color and creates three matching gum balls

= Three matching and 7 ˘ spent

 
Now I had to find out how much a parent would pay if they had three children and a three gum ball machine. It turns out they spend 7 ˘ total to have their children receive three of the same color gum balls.

               

 

 

 

 

 

 

 

 

 

Now that we know how much a parent spends for his children with two colors of gum balls and three colors, what if the parent has five children and five gum ball colors or four children and seven colors? I started by creating my own examples for two children and different number of colors, then three children and a different number of colors and then five children and a different number of colors. This would then create a pattern so it can be applied to any number of children and any number of colors.

  1 ˘      Red

  2 ˘      Blue

  3 ˘     white

  4 ˘     yellow

  5 ˘     one of the colors ( r b y w)

Now the 5th cent is one color and creates two matching gum balls

= two matching and 5 ˘ spent

 
 


Example 1.

                2 children 4 different colors

1,2,3,4  ˘  different colors

5  ˘  one color that will match one of the first four colors.

 

 

 

 

 

  1 ˘      Red

  2 ˘      Blue

  3 ˘     white

  4 ˘     yellow

  5 ˘     pink

  6 ˘    one of the colors ( r b y w p)

Now the 6th cent is one color and creates two matching gum balls

= two matching and 5 ˘ spent

 
Example 2.

                2 children 5 colors

1,2,3,4,5 ˘ different colors

  one color that will match one of the first five colors.

               

 

Example 3.

                3 children 4 colors

1,2,3,4, ˘ different colors

5,6,7,8 ˘ different colors

  1, 2, 3, 4 ˘      Red green blue white

  5, 6, 7, 8 ˘      Red green  Blue white

  9 ˘    one of the colors ( r b y w)

Now the 9th cent is one color and creates three matching gum balls

= three matching and 9 ˘ spent

 
  is one color that matches two from before creating three of the same colors.

 

 

 

 

Example 4.

                3 children 5 gum balls

1,2,3,4,5 ˘ different colors

6,7,8,9,10 ˘ different colors

  1, 2, 3, 4, 5 ˘      Red green blue white pink

   6, 7, 8, 9, 10 ˘      Red green  Blue white pink

 11 ˘    one of the colors ( r b y w p)

Now the 11th cent is one color and creates three matching gum balls

= three matching and 11 ˘ spent

 
11˘  is one color that matches two from before creating three of the same colors.

 

 

 

 

 

Example 5.

                3 children 6 gum balls

1,2,3,4,5,6 ˘ different colors

7,8,9,10,11,12 ˘ different colors

  1, 2, 3, 4, 56  ˘      Red green blue white pink yellow

    7, 8, 9, 10, 11, 12 ˘      Red green  Blue white pink yellow

 13 ˘    one of the colors ( r b g y w p )

Now the 13th cent is one color and creates three matching gum balls

= three matching and 13 ˘ spent

 
13˘  is one color that matches two from before creating three of the same colors.

 

 

 

 

 

Examples 6

                4 children 4 colors

1,2,3,4, ˘ different colors

5,6,7,8 ˘ different colors

9,10,11,12 ˘  different colors

  1, 2, 3, 4 ˘      Red green blue white pink

  5, 6, 7, 8 ˘      Red green  Blue white pink

  9, 10, 11, 12 ˘   

  13 ˘        one of the colors ( r b g w p)

Now the 13th cent is one color and creates four matching gum balls

= three matching and 9 ˘ spent

 
13  ˘ is one color that matches three from before creating four of the same colors.

 

 

 

Examples 6

                4 children 5 colors

1, 2, 3, 4, 5 ˘      Red green blue white pink yellow

6, 7, 8, 9, 10 ˘      Red green  Blue white pink yellow

11, 12, 13, 14, 15 ˘    Red green  Blue white pink yellow

  16 ˘        one of the colors ( r b g w p y)

Now the 16th cent is one color and creates four matching gum balls

= three matching and 9 ˘ spent

 
1,2,3,4,5 ˘ different colors

6,7,8,9,10 ˘ different colors

11,12,13,14,15 ˘  different colors

creating  four of the same colors.

 
16  ˘ is one color that matches three from before

I did this for up to five children and 6 colors (work attached) and I created a chart out of the information.

# children

# colors

 

Costtotal = (G+1)

 
Total cost (˘)

functions

2

2

3 ˘

 

2

3

 

2

4

 

2

5

 

3

3

 

Costtotal = 2G+1

 

 

3

4

 

3

5

11˘

 

3

6

13˘

 

4

4

 

Costtotal = 3G+1

 
13˘

 

4

5

16˘

 

4

6

19˘

 

4

7

 

Costtotal = 4G+1

 
22˘

 

5

5

21˘

 

5

6

25˘

 

5

7

29˘

 

5

8

33˘

 

 

From this I found a pattern in the expressions that I found. To find the total cost for a parent for any given number of children and number of different colored gum balls, the function is the number of children minus one times the number of different colors then plus one. So Costtotal = (C-1)(G)+1

I also used a different method to find this pattern and function and it was more effective and productive. I actually make little charts, almost like the ones in the examples but a little different. The charts showed a patter and reasoning to why adding the gumballs up like I did works.

First 2 children 2 colors Red and Green so first two cents red and green third is a match. This can be written as…..

    R         G             2+1= 3˘

    R

R             G             B

 
Now the same for 2 children 3 colors

                                                3 + 1= 4

R

 
 

 


R             G             B

 
Next is 3 children 3 colors   (3+3) + 1 = 6 + 1 = 7 ˘ also equal to (2 x 3) + 1 = 7 ˘

R             G             B

 

R

 
 


R             G             B             Y

 
3 children 4 colors

R

 

R             G             B             Y

 
                                                                (4 x 2) + 1 = 9 ˘

 

 

R

 

R             G             B             Y

 

R             G             B             Y

 

R             G             B             Y

 
4 children 4 colors

                                                                (4 x 3) + 1= 13 ˘

 

 

 

This pattern shows that you can find the total cost by calculating the number of rows multiplied by the number of columns plus one. And this also shows that to find the total cost for a parent for any given number of children and number of different colored gum balls, the function is the number of children minus one times the number of different colors then plus one. So Costtotal = (C-1)(G)+1. This is a lot simpler than finding individual functions and finding one from there.

And that is how I completed my process.

 

Solution:

                In Sticky Gum Problem we must find a function that will produce the total cost a parent will have to pay if their children want the same color gum ball from a gum call machine. The function must have an input of the number of children and the number of different colors in the gum ball machine. The output must be the total cost in cents to a parent.

                To find the total cost you simply, take the number of children minus one times the number of different colors then plus one. So Costtotal = (C-1)(G)+1.

This is true because if you have two children and three different colors, you will have to use three pennies and receive one of each color and the fourth will no matter what match one of the three previous. This process just repeats with different number of children, if there are 5 children and 5 colors, there will be the first 5 pennies for the first five colors then the next five pennies to make one match then the next five pennies to make a third match and the next five pennies to make a fourth match and then the 21st penny can be any color and it will then create five of the same color gum balls. This process goes for any number of children and number of gumballs. It is just simpler to use the function Costtotal = (C-1)(G)+1. This is called a function because it has two independent variables, two inputs, and produces a dependent variable, the output.

Generalizations:

                Well we know that a function has an input and an output. In our function we know that there will be two inputs, the number of children and the number of different colors. There will only be one output and that is the total cost in cents.

                We also can generalize that when using a gum ball machine you will put in one cent at a time and always receive one gum ball at a time, so there is no cheating the system.

                Also when doing this problem we are only using the function to find the most a parent or any other person would spend. If you are lucky you could receive the same colors immediately in the first few cents and not have to use them all. So the problems are hypothetical situations that can tell a person if their children would like identical color gum balls how much change to be prepared to have.

 

Self-assessment:

                I thought I did well on this POW. It was a lot easier than most POW’s and it only took about one hour to do out on paper. I also tried to prove it with mathematical induction and then realized there were two variables and I couldn’t do it. Hahaha anyway, I thought it was a good POW and I did well on it.

 

By the way…. Are there really gum ball machines with just two or three colors?? J

The answer sure!

http://tbn0.google.com/images?q=tbn:_mD9cpJvpbI1zM:http://www.vendingfriends.com/images/i_gumball_machines.jpg http://tbn0.google.com/images?q=tbn:ARquDJFrdkpJNM:http://rinovelty.com/imagedb/products/_Detail/FRBUBS9.jpg http://tbn0.google.com/images?q=tbn:tykBHlWNt4ZVPM:http://us.st11.yimg.com/us.st.yimg.com/I/igumballs_1974_620768

Haha in appropriate gum ball joke of the day:

A Blonde was at a gumball machine. She put a quarter in and kept getting a gumball out. The man behind her asked if he could get a gumball. She said, "Shut up! I’m WINNING!