A Sticky Gum Problem
This POW was a relatively easy problem and I found the pattern quickly. Here is the sticky gum problem.
Mrs. Hernandez comes across a gum-ball machine one day when she was out with her twins. Of course, the twins each want a gum-ball. They also insist on having gum-balls of the same color. They don't care what color the gum-balls are, as long as they are both the same. There are only two colors of gum-balls in the machine. Since the gum-balls cost a penny each, why is three cents the most she might have to spend?
This is the first question with this problem. The rest of the problem has other variations of this situation, changing the number of kids and the number of colors in the gum-ball machine.
First, I answered all of the questions. For the problem above, the reason is if she gets two of the same color right away, then it costs two cents, if she doesn't, then the next one will have to be one of the colors you already have, giving her two of a kind. The next problem was two kids, and three gum-ball colors. For this problem, if she got one of each (would take three cents so far) the next one will have to give her a color she already has, making the most she might spend four cents.
Next we looked at what would happen if there were triplets and a three color gum-ball machine. This question was a bit harder because there is now another kid. If you have two of each, the next color you get will make a third of each color. The most spent would be seven for this.
The next thing that we did was made up our own scenarios. With four kids, and a three gum-ball machine, it was at most thirteen cents. With five kids, three colors it was at most sixteen and with four kids and two colors it was at most nine cents. The pattern here was pretty simple to discover.
Kids Colors Cents:
2 2 3
2 3 4
3 2 5
3 3 7
3 4 9
4 2 7
4 3 10
So when there are only two kids, the formula is x = (k + c) - 1. X equals max. money spent, K equals the number of kids, and C equals the number of gum-ball colors. For triplets, it is x = (k + c) + (c - 2). Overall, it is x = (k - 1) * c +1.
A general idea that I have is that you should not give in to your kids. Having gum is a privilege and spending five minutes trying to get color coded pieces for triplets is crazy. If it matters that much, go to a drugstore, buy a pack of gum and then all of the kids will have the same gum. I came to this realization when I was working on the second problem, two kids and a three color gum-ball machine. Another thing is that kids should not get gum all the time. In the problem, they are getting gum two days in a row. It is not really good for their teeth. This again was because of the second question again.
Mrs. Hernandez comes across a gum-ball machine one day when she was out with her twins. Of course, the twins each want a gum-ball. They also insist on having gum-balls of the same color. They don't care what color the gum-balls are, as long as they are both the same. There are only two colors of gum-balls in the machine. Since the gum-balls cost a penny each, why is three cents the most she might have to spend?
This is the first question with this problem. The rest of the problem has other variations of this situation, changing the number of kids and the number of colors in the gum-ball machine.
First, I answered all of the questions. For the problem above, the reason is if she gets two of the same color right away, then it costs two cents, if she doesn't, then the next one will have to be one of the colors you already have, giving her two of a kind. The next problem was two kids, and three gum-ball colors. For this problem, if she got one of each (would take three cents so far) the next one will have to give her a color she already has, making the most she might spend four cents.
Next we looked at what would happen if there were triplets and a three color gum-ball machine. This question was a bit harder because there is now another kid. If you have two of each, the next color you get will make a third of each color. The most spent would be seven for this.
The next thing that we did was made up our own scenarios. With four kids, and a three gum-ball machine, it was at most thirteen cents. With five kids, three colors it was at most sixteen and with four kids and two colors it was at most nine cents. The pattern here was pretty simple to discover.
Kids Colors Cents:
2 2 3
2 3 4
3 2 5
3 3 7
3 4 9
4 2 7
4 3 10
So when there are only two kids, the formula is x = (k + c) - 1. X equals max. money spent, K equals the number of kids, and C equals the number of gum-ball colors. For triplets, it is x = (k + c) + (c - 2). Overall, it is x = (k - 1) * c +1.
A general idea that I have is that you should not give in to your kids. Having gum is a privilege and spending five minutes trying to get color coded pieces for triplets is crazy. If it matters that much, go to a drugstore, buy a pack of gum and then all of the kids will have the same gum. I came to this realization when I was working on the second problem, two kids and a three color gum-ball machine. Another thing is that kids should not get gum all the time. In the problem, they are getting gum two days in a row. It is not really good for their teeth. This again was because of the second question again.