In the “Sticky Gum Problem”, we looked at how much money (m) someone might have to spend at a gumball machine with a certain number of colors of gumballs (c), and with a certain number of kids (k) who all want the same color. First, we had to do a bunch of problems that had fixed numbers for k and c, where we had to figure out m, then we had to come up with a generalization (an equation) that worked for all cases of k and c, telling you m.
For the first given problem, we were told about Ms. Hernandez and her twins. They were at a gumball machine with two different colors of gumballs, white (W) and red (R), and the twins each wanted the same color of gumball. The gumballs each cost 1¢. The problem asked why 3¢ was the maximum amount of money Ms. Hernandez would have to spend to get two of one of the colors. I know that she would only have to spend 3¢ because of this logic: we’ll say that first, Ms. Hernandez gets a red gumball (R). (It doesn't matter which you start with, the colors are interchangeable.) Next, she could either get R or W to match up with her R. If she gets R (R, R), she’ll be done with 2¢ spent. If she gets W (R, W), the most expensive possibility, she’ll be set with whatever she gets next, because it will match up with something she already has. Either (R, W, W) or (R, W, R).
For the second problem, Ms. Hernandez and her twins pass a three color gumball machine (with red, white, and blue gumballs), and the kids want the same color again. How much money will Ms. Hernandez have to spend? In the most expensive scenario, Ms. Hernandez would get (R, W B), one of each available color. On her fourth try, she would get either a red (R, W, B, R), white (R, W, B, W) or blue (R, W, B, B), giving her two in whatever color she got last. In all, 4¢ for the most expensive scenario.
In the final problem we did, Mr. Hodges and his triplets pass the three color gumball machine, and the children each want the same color, as before. The maximum amount of money Mr. Hodges would have to spend is found in basically the same way we found the above two answers. The most expensive scenario would be getting the colors in order, putting off getting three of one color for as long as possible (R, W, B, R, W, B). And as above, whatever Mr. Hodges got next would match up with one of the pairs of colors above. Red (R, W, B, R, W, B, R), white (R, W, B, R, W, B, W), or blue (R, W, B, R, W, B, B). This most expensive scenario would cost 7¢ in all, though she could get the gumballs in a less expensive order.
Finally, we had to come up with an equation that would work in any situation, with any number of colors or kids. To attempt to come up with an equation, I looked at what I had done while answering the questions, and played with the variables. I first tried the equation k+(c-1), based on my answers to the first two problems. It worked for the first two problems (with 2 kids), but it didn’t work for the third one.
For the first given problem, we were told about Ms. Hernandez and her twins. They were at a gumball machine with two different colors of gumballs, white (W) and red (R), and the twins each wanted the same color of gumball. The gumballs each cost 1¢. The problem asked why 3¢ was the maximum amount of money Ms. Hernandez would have to spend to get two of one of the colors. I know that she would only have to spend 3¢ because of this logic: we’ll say that first, Ms. Hernandez gets a red gumball (R). (It doesn't matter which you start with, the colors are interchangeable.) Next, she could either get R or W to match up with her R. If she gets R (R, R), she’ll be done with 2¢ spent. If she gets W (R, W), the most expensive possibility, she’ll be set with whatever she gets next, because it will match up with something she already has. Either (R, W, W) or (R, W, R).
For the second problem, Ms. Hernandez and her twins pass a three color gumball machine (with red, white, and blue gumballs), and the kids want the same color again. How much money will Ms. Hernandez have to spend? In the most expensive scenario, Ms. Hernandez would get (R, W B), one of each available color. On her fourth try, she would get either a red (R, W, B, R), white (R, W, B, W) or blue (R, W, B, B), giving her two in whatever color she got last. In all, 4¢ for the most expensive scenario.
In the final problem we did, Mr. Hodges and his triplets pass the three color gumball machine, and the children each want the same color, as before. The maximum amount of money Mr. Hodges would have to spend is found in basically the same way we found the above two answers. The most expensive scenario would be getting the colors in order, putting off getting three of one color for as long as possible (R, W, B, R, W, B). And as above, whatever Mr. Hodges got next would match up with one of the pairs of colors above. Red (R, W, B, R, W, B, R), white (R, W, B, R, W, B, W), or blue (R, W, B, R, W, B, B). This most expensive scenario would cost 7¢ in all, though she could get the gumballs in a less expensive order.
Finally, we had to come up with an equation that would work in any situation, with any number of colors or kids. To attempt to come up with an equation, I looked at what I had done while answering the questions, and played with the variables. I first tried the equation k+(c-1), based on my answers to the first two problems. It worked for the first two problems (with 2 kids), but it didn’t work for the third one.
Before I came up with a second equation, I manually counted (using the third problem as my first example) how many times a gumball was bought, drawing a diagram as an example. Consistently, there was the amount of kids, times the amount of colors, plus the one deciding gumball. My second equation was ck+1, and was based off of this discovery. However, my second equation was incorrect. |
Finally, on my third try, after realizing that I was off by one number with the product of my second equation and the correct answer, and realizing what I had about the actual amounts of things, I came up with c(k-1)+1. If you notice, the only difference between my second and third equations is that I use (k-1) in the third one, and leave k alone in the second. The reason you subtract the amount of kids by one is because one of the kids gets the deciding gumball, which is added in the end, so that kid doesn’t need to also be accounted for when the others are “getting” however many colors there are. The other kids are “getting” however many colors of gumballs, because in the most expensive scenario, the gumballs go in an order so that you don’t get the correct number until the deciding gumball, and that order (for instance (R, W, B)) is repeated (k-1) times.
I know that my equation works, because I tested it on all of the given problems, plus one other one, and it worked for all of them. Here is a data table showing the problems I tested it on:
Kids Color Money
2 2 3¢
2 3 4¢
3 3 7¢
3 4 9¢
While doing this problem, I really stretched my brain. It was really interesting, though difficult, trying to puzzle through the equation (which, in all honesty, I never thought I’d say about finding an equation), and finding it, then understanding it, was really rewarding. I think that I used the Habits of a Mathematician "visualize", and "being confident, patient, and persistent" in this problem. I also used "generalize", by making the equation, but that was actually a part of the problem. I used "visualize" by creating that tiny diagram of the number of colors and kids. It was a small thing, but it brought the whole equation, and later the whole problem, together for me. I was "confident, patient, and mostly persistent" because I used the equation that I knew didn’t work, and made it into the one that I know does work.
I think I deserve a 10/10 for this problem, because of how hard I worked on it. I really did put a lot of effort into it, and I think you can see that in my work.
I know that my equation works, because I tested it on all of the given problems, plus one other one, and it worked for all of them. Here is a data table showing the problems I tested it on:
Kids Color Money
2 2 3¢
2 3 4¢
3 3 7¢
3 4 9¢
While doing this problem, I really stretched my brain. It was really interesting, though difficult, trying to puzzle through the equation (which, in all honesty, I never thought I’d say about finding an equation), and finding it, then understanding it, was really rewarding. I think that I used the Habits of a Mathematician "visualize", and "being confident, patient, and persistent" in this problem. I also used "generalize", by making the equation, but that was actually a part of the problem. I used "visualize" by creating that tiny diagram of the number of colors and kids. It was a small thing, but it brought the whole equation, and later the whole problem, together for me. I was "confident, patient, and mostly persistent" because I used the equation that I knew didn’t work, and made it into the one that I know does work.
I think I deserve a 10/10 for this problem, because of how hard I worked on it. I really did put a lot of effort into it, and I think you can see that in my work.