6/27/13

POW Write-up In this POW write-up, I am trying to prove that there can be only one solution to this problem, and demonstrate and corroborate that all solutions work and are credible. What the problem of the week is asking is that the number that you put in the boxes 0-4 is the number of numbers in the whole 5-digit number. For example, if you put zero in the “one” box, you would be indicating that there is zero ones in the number. Another example is if you put a two in the “three” box. This would indicate that there are two threes in the whole 5-digit number. I was asked to find solutions where are the numbers would work in heir perspective boxes. From there I started working on the problem that would fit this criterion. When I first saw the POW, I thought that finding the solution would be fairly easy, by just plugging different numbers into the boxes and play around with them. I soon discovered that that wasn’t the case. So using deductive reasoning, I began off by acknowledging that there couldn’t be any number over four in the boxes, because if it was above four, it wouldn’t be the right solution. If I put a five in any box, I knew that there would only be four boxes left, which wouldn’t work. So I ruled out any number over four. From there, I began by using the largest number box, which was four. I noticed that if I used any number other than zero in the four box, the problem wouldn’t work. To prove this, if I put a one in the “four” box, that means that there would be another box from 0-3 that would contain the number four, which means that there would be four of that number. This is impossible because there are only three boxes left. Any other solution wouldn’t work either, so I left the “four” box as a zero. Then I went down to the “three” box, being quite confident of my decision on the “four” box. I used the same approach, and found out that there couldn’t be any number over two, because if you put a three in the “three” box, there would have to be three more three’s in the rest of the three boxes, which wouldn’t make sense because there would be no space to fit the three zeros, ones, or twos. If I put a two, then that means that there would be two threes in the rest of the boxes, which wouldn’t work either. I tried plugging in a one in the “three” box, but that wouldn’t work either because then there would be one three in any of the boxes, which wouldn’t be fulfilled. So the only solution left would be a zero in the “three” box. Moving on to the last boxes, I figured out that the only numbers to work with would be zero, one, and two. I noticed that in the “zero” box, there were already two zeros in play, so the number in the “zero” box had to be two or above. I tried plugging in a three, but that wouldn’t work out because there is already a zero in the “three” box. So I left it as a two. From there on, I had two numbers left to work with (hopefully). I used my reasoning to find out that there was a two in the “zero” box, which meant that there had to be at least one “one” in the two box. I tried that, and I found out that if I put a one in the “two” box, I wouldn’t have any correct solution for the “one” box. If I put a one in the “one” box, there would be two ones, which would contradict the solution I was trying to find. If I put a two in the “two” box, then it would be perfect, because then there would be two twos; one in the “zero” box and another in the “two” box. So then I was left alone with the “one” box, where I conveniently found out I could put a one in the “one” box, and it would work. The answer to the POW is the number 21200. I know this is the only solution because our geometry teacher, Mr. Carter (or as he likes us to call him, Dave), told our class so. Starting over, I know that there could only be a zero in the “four” box, because any other number wouldn’t work in the four box, because there wouldn’t be enough boxes to fulfill the problem. I am certainly definite that there could only be a zero in the “four” box. As I said before, there could only be numbers 4 and under. Moving onto the “three” box, there could only be a zero because if you put a one in the box, there would eventually have to be a three somewhere in the other three boxes, which wouldn’t work because there wouldn’t have space to fit those because there would only be two boxes left. So moving on to the “zero” box, there would have to be a number two or above. If there was a three in the “zero” box, then that would mean there would have to be a zero in the “one” or “two” box, which wouldn’t add up because one or the other wouldn’t have a right solution. So the answer to that would be a “two” to make up for the “four” and “five” boxes. Being left with the “one” and “two” boxes, I noticed that since there was a two in the “zero” box, there would be at least a one in the “two” box. I tried that, but then the “one” box would have to contain a two, which wouldn’t work. Therefore, the “two” box has to contain a two. In the “one” box, there could only be a one to show that there is only one “one”. I know this is the only solution because I used deductive reasoning and slowly eliminated choices and came down to only one answer. Therefore I stand by my conclusion that the only solution is the number 21200. For this POW write-up, I think I deserve an A/A- because I spent a lot of effort and time into my POW write-up and tried to defend my thinking as well as possible. I tried to make the reader see what I was seeing and look into my logic and my thinking. I also think I completed the rubric to the best of my ability and did all the steps that were asked for. I used a step-by-step process to get to my solution. I also mentioned some problems and difficulties I faced while solving the POW. Therefore, I think I am worthy of receiving an A/A- for the POW write-up. Thank you!