Algebra 1,
Pow Exhibition April 29th:
For this exhibition I presented the ninth problem of the week that we did this year. This POW is called, " Colored Cubes" and deals with a 5x5 cube that is made up of smaller cubes. Each smaller cube is painted on all exposed faces, the question of the POW is, " How many cubes have 1 face painted, 2 faces painted, and 3 faces painted." To prepare this POW for presenting I wrote down everything I knew about it and looked for spots that I could improve my knowledge on. For this POW that was the equations, to combat my inexperience with the equations I practiced using them with a multitude of different numbers. After this I just rote all of my thoughts down and presented that. Teaching this Pow helped improve my understanding of the equations (N-2)2 -6 and 12(N-2).
March 28th POW 10:
POW 10
By Dylan King
Process
In the10th pow students were required to find an answer for the following question,” A camel has a banana farm that yields 3000 bananas, she want’s to sell the banana’s at a market that is 1000 miles away. However, the camel can only carry 1000 of the 3000 bananas at a time, the camel also eats 1 banana for every mile it walks.” At first, this problem seems impossible but upon closer inspection I realized that the camel could leave any number of bananas laying along it’s path, as long as it has enough to get back. Now that I understood this I knew the problem would be easier, but before tackling this bigger problem I decided to complete the mini POW first.
Similar to the larger POW the mini POW has a camel that yields 45 bananas and sells them at a market 15 miles away, the problem is she can only carry 15 bananas at a time and eats a banana every mile it walks. To start this problem I went to the whiteboard because I knew that this POW would require lot’s of erasing and re-doing. I first drew a line across the middle of the whiteboard this would symbolize the 15 miles that the camel had to walk to get to the market. On my first attempt I had the camel walk five miles and drop off five of the 15 bananas there, I repeated this once more until there were only 15 bananas left out of the original 45. After i did this I had the camel walk 5 miles and pick up the 5 bananas, I believed I had cracked the code and gotten fifteen bananas to the market , but then I realized that the camel would need to eat 10 bananas on the way to the market which would leave it with only 5 bananas to sell at the market. Although my first try failed my second attempt gave me some perspective on the mini POW although I did not get the correct answer. In my second try I had the camel drop 5 bananas off at the 5 mile mark, then I had it drop off one banana at the 7 mile mark. This meant that on the final trip the camel was able to make it to the market with a total of 6 bananas. The only reason this worked is because anything over seven miles would mean that the camel could not return to pick up the rest of it’s bananas,and if I had dropped of two bananas at the 6 mile mark one banana would have been wasted. After a week working on this and evaluating the problem with Lauren, I figured out a way to get more bananas to the market. If you drop 18 bananas at the three mile mark over two trips then you would be able to take 15 bananas to the eight mile mark then the camel can go back to the three mile mark and pick up the last fifteen. This means that the camel can get eight bananas to the market without wasting any bananas. Now that I had some idea of what I was doing I moved on to the big problem in which the camel has 1000 miles to traverse. I used the same method that I did in the mini POW which enabled me to get a good answer, the first thing I did was have the camel walk 200 miles and drop off 600 bananas I repeated this twice. That means that there would be 1,200 bananas waiting at the 200 mile mark on the camels last trip. The end result to this was the camel being able to get 530 bananas to the market. It was at this point that I realized that both of these problems are the same, the big pow is the blown up version of the mini POW. All in all, thinking through both of these problems and taking my time has allowed me to get 53% of the number of bananas that the camel could carry to the market, resulting in profit for the camel.
Solution
The answer to the 10th POW is that the camel can get 530 bananas to the market , as explained in the process the camel leaves 1,200 bananas at the 200 mile mark and ends up with 530 bananas to take to the market. 530 is the right answer because if you go over the 200 mile strategy you either wouldn’t be able to leave any bananas and if you go under 200 you are wasting bananas and through all of my attempts I was never able to get anything over 530 bananas to the market. As for the mini POW I used the same kind of process and got 8 of the bananas to the market. To get 8 bananas to the market the camel has to drop 18 bananas at the 3 mile mark and then 5 bananas at the 8 mile mark, this ensures that the camel can get 8 bananas to the market. This is the right answer because you can’t go over seven and go back successfully and if you stay under seven you would waste bananas. Through all of my attempts with the mini POW I was never able to get any more than 6 bananas to the market. In conclusion, this is the solution to the tenth POW.
Extension
For this problem a hard extension would make it so the camel had a limit of miles it could walk in order to get the bananas to the market. This would vastly increase the difficulty of this problem because the camel would not be able to go back and forth freely as I explained in my problem, this would make it almost impossible to get a figure like 400 or 6 to the market.An easier extension to this problem would have the camel able to carryup to 2000 or 30 bananas unlike the 1000 or 15 current constraint. All in all, these are some examples of extensions for the tenth POW.
Evaluation
I believe that this is one of the most interactive and creative POW’s so far this year and I liked it the most out of all of the POW’s so far this semester. I liked the process of creating the drawings and following through with this POW. My least favorite part of this POW is definitely the mini POW in addition to the bigger POW, although this is a minor complaint I would have liked to have the mini POW due on Wednesday and be able to finish the big question over spring break. Other than that minor complaint I think that this was my second favorite POW overall. In conclusion, this is my evaluation of the tenth POW.
By Dylan King
Process
In the10th pow students were required to find an answer for the following question,” A camel has a banana farm that yields 3000 bananas, she want’s to sell the banana’s at a market that is 1000 miles away. However, the camel can only carry 1000 of the 3000 bananas at a time, the camel also eats 1 banana for every mile it walks.” At first, this problem seems impossible but upon closer inspection I realized that the camel could leave any number of bananas laying along it’s path, as long as it has enough to get back. Now that I understood this I knew the problem would be easier, but before tackling this bigger problem I decided to complete the mini POW first.
Similar to the larger POW the mini POW has a camel that yields 45 bananas and sells them at a market 15 miles away, the problem is she can only carry 15 bananas at a time and eats a banana every mile it walks. To start this problem I went to the whiteboard because I knew that this POW would require lot’s of erasing and re-doing. I first drew a line across the middle of the whiteboard this would symbolize the 15 miles that the camel had to walk to get to the market. On my first attempt I had the camel walk five miles and drop off five of the 15 bananas there, I repeated this once more until there were only 15 bananas left out of the original 45. After i did this I had the camel walk 5 miles and pick up the 5 bananas, I believed I had cracked the code and gotten fifteen bananas to the market , but then I realized that the camel would need to eat 10 bananas on the way to the market which would leave it with only 5 bananas to sell at the market. Although my first try failed my second attempt gave me some perspective on the mini POW although I did not get the correct answer. In my second try I had the camel drop 5 bananas off at the 5 mile mark, then I had it drop off one banana at the 7 mile mark. This meant that on the final trip the camel was able to make it to the market with a total of 6 bananas. The only reason this worked is because anything over seven miles would mean that the camel could not return to pick up the rest of it’s bananas,and if I had dropped of two bananas at the 6 mile mark one banana would have been wasted. After a week working on this and evaluating the problem with Lauren, I figured out a way to get more bananas to the market. If you drop 18 bananas at the three mile mark over two trips then you would be able to take 15 bananas to the eight mile mark then the camel can go back to the three mile mark and pick up the last fifteen. This means that the camel can get eight bananas to the market without wasting any bananas. Now that I had some idea of what I was doing I moved on to the big problem in which the camel has 1000 miles to traverse. I used the same method that I did in the mini POW which enabled me to get a good answer, the first thing I did was have the camel walk 200 miles and drop off 600 bananas I repeated this twice. That means that there would be 1,200 bananas waiting at the 200 mile mark on the camels last trip. The end result to this was the camel being able to get 530 bananas to the market. It was at this point that I realized that both of these problems are the same, the big pow is the blown up version of the mini POW. All in all, thinking through both of these problems and taking my time has allowed me to get 53% of the number of bananas that the camel could carry to the market, resulting in profit for the camel.
Solution
The answer to the 10th POW is that the camel can get 530 bananas to the market , as explained in the process the camel leaves 1,200 bananas at the 200 mile mark and ends up with 530 bananas to take to the market. 530 is the right answer because if you go over the 200 mile strategy you either wouldn’t be able to leave any bananas and if you go under 200 you are wasting bananas and through all of my attempts I was never able to get anything over 530 bananas to the market. As for the mini POW I used the same kind of process and got 8 of the bananas to the market. To get 8 bananas to the market the camel has to drop 18 bananas at the 3 mile mark and then 5 bananas at the 8 mile mark, this ensures that the camel can get 8 bananas to the market. This is the right answer because you can’t go over seven and go back successfully and if you stay under seven you would waste bananas. Through all of my attempts with the mini POW I was never able to get any more than 6 bananas to the market. In conclusion, this is the solution to the tenth POW.
Extension
For this problem a hard extension would make it so the camel had a limit of miles it could walk in order to get the bananas to the market. This would vastly increase the difficulty of this problem because the camel would not be able to go back and forth freely as I explained in my problem, this would make it almost impossible to get a figure like 400 or 6 to the market.An easier extension to this problem would have the camel able to carryup to 2000 or 30 bananas unlike the 1000 or 15 current constraint. All in all, these are some examples of extensions for the tenth POW.
Evaluation
I believe that this is one of the most interactive and creative POW’s so far this year and I liked it the most out of all of the POW’s so far this semester. I liked the process of creating the drawings and following through with this POW. My least favorite part of this POW is definitely the mini POW in addition to the bigger POW, although this is a minor complaint I would have liked to have the mini POW due on Wednesday and be able to finish the big question over spring break. Other than that minor complaint I think that this was my second favorite POW overall. In conclusion, this is my evaluation of the tenth POW.
Pow #9
POW 9
By Dylan King
Process
In this Problem of the Week students were tasked with deciphering a rubix cube in order to determine the number of cubes that would be painted and how many would have corresponding colors. To begin this problem I attempted drawing a 5x5 cube that I could simulate this problem with, after wasting fifteen minutes I decided that the picture of a rubix cube that came along with the POW would do. To answer the first question, how many small cubes have only one face painted, I decided to count the number of cubes on one side and then multiply that by six which is the total number of sides on a cube. During the process of solving this problem I discovered that the outer rim of smaller cubes would always have two or three faces painted which made it easier to spot the cubes with only one face painted. Afterwards I decided to move to the third question which asked how many cubes would have three faces painted, this was relatively easy to solve because there were only eight possible places where a cube could be painted on three sides. When I had completed this I returned to the question of which cubes would have two faces painted. This was the most challenging question for me because I tried to count all of the cubes one by one and that ended up wasting a lot of my time and getting an incorrect answer. Although I knew it would be a multiple of twelve I had trouble trying to count the side that I could not see and in the end that was where I lost count and got a wrong answer; after many confused hours I figured out the correct answer. After this I moved on to looking for patterns and generalities, I realized that no matter the size of the cube,(NxN) there will always only be eight cubes that have three sides painted.All in all this is the process that I took to understanding the 9th POW
Solution
This Problem of The Week asks a plethora of questions concerning a cube and the smaller cubes that go into it. My answer to the first question of this POW is 54, I got this answer from counting the number of small cubes with only one face painted on one side of the rubix cube. This gave me a total of nine, then I multiplied nine by the number of sides on a rubix cube which is 6, this gave me the answer which I believe to be 54. Next I looked at the second question that we are presented with, how many cubes have two faces painted, as I explained above this took me quite some time, but I now know that the answer is 36. I got 36 by counting all of the cubes that had two faces painted on one side of the rubix cube, this gave me twelve , after that I began to count the rest leaving out the sides and cubes that I had already counted this gave me a grand total of 36 cubes that have two faces painted. Finally, I looked at all of the cubes that had three faces painted, for me this was the easiest problem to solve because I could see that the only time a cube had three faces painted was on the edge of the rubix cube. With there only being 8 edges on a rubix cube I was able to figure out that the total number of cubes with three faces painted was 8. This left me with 27 cubes that were not painted and no cubes that were painted on 4 sides.
3 sides painted
2 sides painted
1 side painted
0 sides painted
8
36
54
27
After working on this problem for a week I have concluded that there is no possible combination of NxNxN cube that would give an example of cubes painted on 4 sides, 5 sides ,and 6 sides. In conclusion, these are my answers to the 9th POW.
Extension
To make this Problem of the Week harder I would ask for the total number of cubes in a 20x20x20 cube and for the student to find out how many cubes would not have any paint on them. In contrast for an easier route I would have used a 3x3 cube which would make this problem far less challenging. In the end, these are my suggested extensions for this Problem of the Week.
Evaluation
In my opinion this Problem of the Week was executed near perfectly, I thought that the difficulty was well balanced and that we had enough time to complete it. Although we did not work on this in class I thought that this was a relatively straightforward problem and did not require class time to understand. The one thing that I would do regarding this problem and any future ones would be to go over the core mechanics and equations that go into the POW after it is due. I believe that this would help me and anyone else with their understanding of the POW. In conclusion these are my thoughts and reflections of this POW.
By Dylan King
Process
In this Problem of the Week students were tasked with deciphering a rubix cube in order to determine the number of cubes that would be painted and how many would have corresponding colors. To begin this problem I attempted drawing a 5x5 cube that I could simulate this problem with, after wasting fifteen minutes I decided that the picture of a rubix cube that came along with the POW would do. To answer the first question, how many small cubes have only one face painted, I decided to count the number of cubes on one side and then multiply that by six which is the total number of sides on a cube. During the process of solving this problem I discovered that the outer rim of smaller cubes would always have two or three faces painted which made it easier to spot the cubes with only one face painted. Afterwards I decided to move to the third question which asked how many cubes would have three faces painted, this was relatively easy to solve because there were only eight possible places where a cube could be painted on three sides. When I had completed this I returned to the question of which cubes would have two faces painted. This was the most challenging question for me because I tried to count all of the cubes one by one and that ended up wasting a lot of my time and getting an incorrect answer. Although I knew it would be a multiple of twelve I had trouble trying to count the side that I could not see and in the end that was where I lost count and got a wrong answer; after many confused hours I figured out the correct answer. After this I moved on to looking for patterns and generalities, I realized that no matter the size of the cube,(NxN) there will always only be eight cubes that have three sides painted.All in all this is the process that I took to understanding the 9th POW
Solution
This Problem of The Week asks a plethora of questions concerning a cube and the smaller cubes that go into it. My answer to the first question of this POW is 54, I got this answer from counting the number of small cubes with only one face painted on one side of the rubix cube. This gave me a total of nine, then I multiplied nine by the number of sides on a rubix cube which is 6, this gave me the answer which I believe to be 54. Next I looked at the second question that we are presented with, how many cubes have two faces painted, as I explained above this took me quite some time, but I now know that the answer is 36. I got 36 by counting all of the cubes that had two faces painted on one side of the rubix cube, this gave me twelve , after that I began to count the rest leaving out the sides and cubes that I had already counted this gave me a grand total of 36 cubes that have two faces painted. Finally, I looked at all of the cubes that had three faces painted, for me this was the easiest problem to solve because I could see that the only time a cube had three faces painted was on the edge of the rubix cube. With there only being 8 edges on a rubix cube I was able to figure out that the total number of cubes with three faces painted was 8. This left me with 27 cubes that were not painted and no cubes that were painted on 4 sides.
3 sides painted
2 sides painted
1 side painted
0 sides painted
8
36
54
27
After working on this problem for a week I have concluded that there is no possible combination of NxNxN cube that would give an example of cubes painted on 4 sides, 5 sides ,and 6 sides. In conclusion, these are my answers to the 9th POW.
Extension
To make this Problem of the Week harder I would ask for the total number of cubes in a 20x20x20 cube and for the student to find out how many cubes would not have any paint on them. In contrast for an easier route I would have used a 3x3 cube which would make this problem far less challenging. In the end, these are my suggested extensions for this Problem of the Week.
Evaluation
In my opinion this Problem of the Week was executed near perfectly, I thought that the difficulty was well balanced and that we had enough time to complete it. Although we did not work on this in class I thought that this was a relatively straightforward problem and did not require class time to understand. The one thing that I would do regarding this problem and any future ones would be to go over the core mechanics and equations that go into the POW after it is due. I believe that this would help me and anyone else with their understanding of the POW. In conclusion these are my thoughts and reflections of this POW.
February 10th POW 8:
Pow 8
By Dylan King
To accomplish this POW students had to delve into a dilemma that a king is faced with when he equally spreads all of the gold in the kingdom between 8 different people. He gets world that one of these 8 have been using the gold for their own selfish purposes. He sends for all 8 people and uses a balance scale to weigh each bag of gold, the problem is that the king wants the scale to be used the least amount of times possible. He thinks that three weighings are the minimum amount of times needs to find the thief, however the court mathematician says it can be done in less. The goal for students was to figure out the least possible uses of the scale in order to find the stealer. I started this problem out just like everyone else, drawing scales and circles all over a sheet of paper until I was very confused and agitated. I began by making a scale that had 4 bags of money on each side and eventually I began to think outside of the box which lead to an inevitable answer.
The least number of weighings that need be performed in order to solve this problem is, 2. I found this by leaving two of the 8 bags out while I weighed the other six, with 3 on each side of the scale. If the 3 bags on each side are even and do not alter in weight then one of the two left out is the theifs and by weighing those 2 together you can get an answer in only 2 weighings. On the other hand if the scale is uneven while weighing the 3 bags on each side you can take the lighter 3 bags and leave one out while you weigh the two others. Both of these solutions will only take two weighings to complete ,thus solving the problem.
An extension that I would propose for this problem would be to increase the bags and require less than ten weighings on the scale. This would make the problem very difficult, even more so than it already is. All in all, this is a challenging extension for the 8th POW.
The only feedback I could offer for this Pow would be to spend more class time working on the core of the problem. Other then that I believe this POW was successful in being semi-fun without being too hard or too easy. For me this is one of the harder POW’s we have done all year, besides that I enjoyed working on it. In the end, this is my feedback to for the 8th Pow.
By Dylan King
To accomplish this POW students had to delve into a dilemma that a king is faced with when he equally spreads all of the gold in the kingdom between 8 different people. He gets world that one of these 8 have been using the gold for their own selfish purposes. He sends for all 8 people and uses a balance scale to weigh each bag of gold, the problem is that the king wants the scale to be used the least amount of times possible. He thinks that three weighings are the minimum amount of times needs to find the thief, however the court mathematician says it can be done in less. The goal for students was to figure out the least possible uses of the scale in order to find the stealer. I started this problem out just like everyone else, drawing scales and circles all over a sheet of paper until I was very confused and agitated. I began by making a scale that had 4 bags of money on each side and eventually I began to think outside of the box which lead to an inevitable answer.
The least number of weighings that need be performed in order to solve this problem is, 2. I found this by leaving two of the 8 bags out while I weighed the other six, with 3 on each side of the scale. If the 3 bags on each side are even and do not alter in weight then one of the two left out is the theifs and by weighing those 2 together you can get an answer in only 2 weighings. On the other hand if the scale is uneven while weighing the 3 bags on each side you can take the lighter 3 bags and leave one out while you weigh the two others. Both of these solutions will only take two weighings to complete ,thus solving the problem.
An extension that I would propose for this problem would be to increase the bags and require less than ten weighings on the scale. This would make the problem very difficult, even more so than it already is. All in all, this is a challenging extension for the 8th POW.
The only feedback I could offer for this Pow would be to spend more class time working on the core of the problem. Other then that I believe this POW was successful in being semi-fun without being too hard or too easy. For me this is one of the harder POW’s we have done all year, besides that I enjoyed working on it. In the end, this is my feedback to for the 8th Pow.
Pow #7:
POW 7
By Dylan King
In this Problem of The Week,(POW), students were asked to, “ Imagine that you have a large number of mechanical clocks in your home (these are traditional analog non digital 12 hour clocks.) Unfortunately, none your clocks keeps time properly. Each clock consistently gains or loses a fixed number of minutes each hour. But different clocks may gain or lose different amounts. At noon today, all the clocks are correct.” Using a chart I was able to start on a simple variant of the required 3,7, and 10 minute changes in time each hour. To get an understanding of this problem the entire class worked out what a 30 minute gain each hour would equate to; after working for almost an hour we found that it took 24 hours for the clocks to match up again. At this point I still had a limited understanding of how the POW worked, I decided to continue with the chart writing an example of which is:
This way took up an absurd amount of time and lead on to me wasting 14 pieces of paper ,front and back, trying to solve the 3 main questions which are:
If a clock gains 10 minutes each hour ,when will it be correct again?
If a clock loses 3 minutes each hour, when will it be correct again
If a clock gains 7 minutes each hour, when will it be correct again
To figure out the correct answer I had one column which displayed the time in regular 12 hour periods and in the other I had the different clock times (refer to example above.) Although this took an extremely long time I eventually solved the seventh POW. First I tried the 10 minute problem which took about half an hour and revealed that it takes 72 hours for the regular 12 hour clock to match up with the clock that adds ten minutes each hour. Next, I worked on tackling the three minute subtraction problem, this problem took the most amount of time and paper. Eventually I found that it takes 240 hours for the clocks to align. While I was solving this problem I discovered a pattern that happens when the out of order clock strikes an eve hour, the first time the odd clock is an hour behind the regular clock; the second time the odd clock is two hours behind the regular clock. After the odd clock is 11 hours behind the regular clock the next time the odd clock hits an even hour you will have your answer, which in this case is 240 hours. After all of the time I put into this problem this was the only available pattern that I could see. All in all, this is how I went about solving the seventh POW.
A challenge extension for this problem would be to alter the original time which means the challenge extension will have a 13 to 23 hour clock as opposed to a regular twelve hour clock. This would greatly increase the difficulty of the problem. All in all, this is what I would consider a difficult challenge extension to this problem of the week.
Reflecting on this problem I would have to say that as daunting as it first appeared this problem was an easily accomplishable medium difficulty compared to some of our other POW’s. This problem was a good mix of difficulty and time that made working on it more fun than the last two more rushed POW’s. If I had any suggestions to make they would be to present this problem with a little more context in this area, but of course that is just my personal opinion. In the end, this was a fun problem and I had an enjoyable time working on it.
In conclusion, these have been my findings and thoughts on the seventh Problem of The Week.
By Dylan King
In this Problem of The Week,(POW), students were asked to, “ Imagine that you have a large number of mechanical clocks in your home (these are traditional analog non digital 12 hour clocks.) Unfortunately, none your clocks keeps time properly. Each clock consistently gains or loses a fixed number of minutes each hour. But different clocks may gain or lose different amounts. At noon today, all the clocks are correct.” Using a chart I was able to start on a simple variant of the required 3,7, and 10 minute changes in time each hour. To get an understanding of this problem the entire class worked out what a 30 minute gain each hour would equate to; after working for almost an hour we found that it took 24 hours for the clocks to match up again. At this point I still had a limited understanding of how the POW worked, I decided to continue with the chart writing an example of which is:
This way took up an absurd amount of time and lead on to me wasting 14 pieces of paper ,front and back, trying to solve the 3 main questions which are:
If a clock gains 10 minutes each hour ,when will it be correct again?
If a clock loses 3 minutes each hour, when will it be correct again
If a clock gains 7 minutes each hour, when will it be correct again
To figure out the correct answer I had one column which displayed the time in regular 12 hour periods and in the other I had the different clock times (refer to example above.) Although this took an extremely long time I eventually solved the seventh POW. First I tried the 10 minute problem which took about half an hour and revealed that it takes 72 hours for the regular 12 hour clock to match up with the clock that adds ten minutes each hour. Next, I worked on tackling the three minute subtraction problem, this problem took the most amount of time and paper. Eventually I found that it takes 240 hours for the clocks to align. While I was solving this problem I discovered a pattern that happens when the out of order clock strikes an eve hour, the first time the odd clock is an hour behind the regular clock; the second time the odd clock is two hours behind the regular clock. After the odd clock is 11 hours behind the regular clock the next time the odd clock hits an even hour you will have your answer, which in this case is 240 hours. After all of the time I put into this problem this was the only available pattern that I could see. All in all, this is how I went about solving the seventh POW.
A challenge extension for this problem would be to alter the original time which means the challenge extension will have a 13 to 23 hour clock as opposed to a regular twelve hour clock. This would greatly increase the difficulty of the problem. All in all, this is what I would consider a difficult challenge extension to this problem of the week.
Reflecting on this problem I would have to say that as daunting as it first appeared this problem was an easily accomplishable medium difficulty compared to some of our other POW’s. This problem was a good mix of difficulty and time that made working on it more fun than the last two more rushed POW’s. If I had any suggestions to make they would be to present this problem with a little more context in this area, but of course that is just my personal opinion. In the end, this was a fun problem and I had an enjoyable time working on it.
In conclusion, these have been my findings and thoughts on the seventh Problem of The Week.
January 20th POW 6:
POW 6
By Dylan King
In this, “Problem of the Week” (pow) students were tasked with examining a phone tree problem. The constraints of this problem stated that , “ Charlotte” had one hour to call her friends to tell them about important news. In this case Charlotte had figured out that each call would be three minutes long in order to convey her intended message. In this instance Charlotte can make twenty calls in her allotted time the first of which she makes to Mike and in turn he has the time to call 19 more people in the hour, and so this pattern continues for the hour thus creating a phone tree.
Process:At first I thought this problem was simple and could be solved in less than an hour. My first reaction to this problem was that if I layed it all out the answer would present itself. I thought perhaps the answer was 210 since that is the sum of 1 person calling 20, the next calling 18, the next calling 17, etc. My dad suggested I use excel to lay out the scenarios. I then promptly put the data into a microsoft excel spreadsheet typing two columns one representing the number of people taking part in the phone tree and the other represented the number of people called and who they could call
person
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
people called
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
The answer from excel was that Charlotte and her phone tree were able to reach 210 people. That seemed too simple and it did not involve exponents. After deeper research into this problem I realized that this was the incorrect answer and continued my efforts to resolve this POW.
Solution: I started by working this problem backwards. An hour has 60 minutes and each phone call takes 3 minutes, so there would be 20 phone calls that each person can make. One person starts by calling another person ,then those two people can call two others. Those four people can call two people each I saw that the number of people calling was doubling each time and this is how I came up with an answer. The answer to the 6th Problem of the Week is 1,048,576. I’ve reached this answer through the process of having two people per line in the phone tree and multiplying that by twenty, which is the amount of calls that Charlotte can make in an hour and in the end I figured out that 2 to the 20th power was the exponent I was searching for. In conclusion, this is how I figured out the 6th Problem of the Week.
Extension: A harder extension to this problem would be to say that every other call lasted 5 minutes instead of the regular three. This would cause the phone tree problem to become very difficult and an answer almost impossible to find. Contrasting this route, is the easier version of this problem which could be that everyone in the phone tree only contacted 5 other people or only had five other friends. This would greatly decrease the difficulty of the 6th POW.
Evaluation: Even though this problem had its difficulties I believe it was a worthwhile problem to pursue and lead to a lot of struggle while working to find an answer. If I had had anything that I would change about this problem it would be to give us more class time to complete it. Although, I did enjoy the toughness that came along with this problem and working on it with my parents and classmates was very fun. As far as the difficulty I feel that this problem was very difficult and challenging and I think that it was a good thing to mix this in with some of the other medium difficulty POW’s that we have done in the past. All in all, this was a fun way to work on a Problem of the Week and I’m looking forward to many more similar ones.
By Dylan King
In this, “Problem of the Week” (pow) students were tasked with examining a phone tree problem. The constraints of this problem stated that , “ Charlotte” had one hour to call her friends to tell them about important news. In this case Charlotte had figured out that each call would be three minutes long in order to convey her intended message. In this instance Charlotte can make twenty calls in her allotted time the first of which she makes to Mike and in turn he has the time to call 19 more people in the hour, and so this pattern continues for the hour thus creating a phone tree.
Process:At first I thought this problem was simple and could be solved in less than an hour. My first reaction to this problem was that if I layed it all out the answer would present itself. I thought perhaps the answer was 210 since that is the sum of 1 person calling 20, the next calling 18, the next calling 17, etc. My dad suggested I use excel to lay out the scenarios. I then promptly put the data into a microsoft excel spreadsheet typing two columns one representing the number of people taking part in the phone tree and the other represented the number of people called and who they could call
person
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
people called
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
The answer from excel was that Charlotte and her phone tree were able to reach 210 people. That seemed too simple and it did not involve exponents. After deeper research into this problem I realized that this was the incorrect answer and continued my efforts to resolve this POW.
Solution: I started by working this problem backwards. An hour has 60 minutes and each phone call takes 3 minutes, so there would be 20 phone calls that each person can make. One person starts by calling another person ,then those two people can call two others. Those four people can call two people each I saw that the number of people calling was doubling each time and this is how I came up with an answer. The answer to the 6th Problem of the Week is 1,048,576. I’ve reached this answer through the process of having two people per line in the phone tree and multiplying that by twenty, which is the amount of calls that Charlotte can make in an hour and in the end I figured out that 2 to the 20th power was the exponent I was searching for. In conclusion, this is how I figured out the 6th Problem of the Week.
Extension: A harder extension to this problem would be to say that every other call lasted 5 minutes instead of the regular three. This would cause the phone tree problem to become very difficult and an answer almost impossible to find. Contrasting this route, is the easier version of this problem which could be that everyone in the phone tree only contacted 5 other people or only had five other friends. This would greatly decrease the difficulty of the 6th POW.
Evaluation: Even though this problem had its difficulties I believe it was a worthwhile problem to pursue and lead to a lot of struggle while working to find an answer. If I had had anything that I would change about this problem it would be to give us more class time to complete it. Although, I did enjoy the toughness that came along with this problem and working on it with my parents and classmates was very fun. As far as the difficulty I feel that this problem was very difficult and challenging and I think that it was a good thing to mix this in with some of the other medium difficulty POW’s that we have done in the past. All in all, this was a fun way to work on a Problem of the Week and I’m looking forward to many more similar ones.
December 14th POW 5:
POW #5
By Dylan King
In POW 5,” Bales and Bales of Hay” students are tasked with taking the weight of multiple bales of hay measured all at once and finding the weight of the individual bales of hay.
to simulate this problem I started off using:
A B C D E
to represent the weight of each bale of hay.
To start this problem off let’s look at the information given to us via the instruction sheet for, “Bales and Bales of Hay”
“The weight of these combinations were written down and arranged in numerical order, the weights are: 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91.”
Now that we know this we can start to visualize the way in which we can obtain the weight of the single bales of hay.After many frustrating tries I realized that the only way to get the correct answer was to use the system below to fill in the blanks.
A+B=80
A+C=82
A+D=83
A+E=84
B+C=85
B+D=86
B+E=87
C+D=88
C+E=90
D+E=91
After discovering this system I plugged in various numbers such as 31, 42, 49, and even 50 the reason that I did this is because our teacher advised us to start in the thirties and not the twenties or lower. I kept trying different possibilities until I got a solution, and in this problems case the solution is:
39+41=80
39+43=82
39+44=83
39+45=84
41+44=85
41+45=86
41+46=87
43+45=88
43+47=90
44+47=91
The numbers are not as orderly as I wished for them to be. It is odd because the system I used to formulate the right answer yields incorrect numbers when I add them all together. Maybe this stroke of luck just saved me three points towards this assignment.
This is the correct way to solve this problem because each two weights add to get the initial group weight. For example,
39+41=80
So on and so forth this goes until we reach:
44+47=91
At this point all of our answers have been solved and we can come to a conclusion for this problem. As far as I know there are no other ways to solve this problem and if there where it seems like they would be harder than the simple adding used in my method of answering this problem. All in all, this is the process I followed to reach my answer to the, “Bales and Bales of Hay” problem.
An extension to this problem could be making the number of bales continue and deviate from the original pattern of solving for the weight. For example you would have to convert bales measured in kilograms to pounds and those measured in pounds to kilograms. A challenging addition to this problem would be to combine decimals and fractions with the already difficult problem. In Conclusion, these are two ways in which the problem could become more challenging and interesting.
In my opinion this problem was a good way to spend our class time, it was challenging enough to leave us stumped for a few days but still easy enough to come up with a clean cut answer to. I thought this problem was relevant and important in bettering our problem solving skills.I wouldn’t do anything to change this problem or others like it and I enjoyed working on it in class. In conclusion, this is my personal reflection of POW 5, “Bales and Bales of Hay.”
By Dylan King
In POW 5,” Bales and Bales of Hay” students are tasked with taking the weight of multiple bales of hay measured all at once and finding the weight of the individual bales of hay.
to simulate this problem I started off using:
A B C D E
to represent the weight of each bale of hay.
To start this problem off let’s look at the information given to us via the instruction sheet for, “Bales and Bales of Hay”
“The weight of these combinations were written down and arranged in numerical order, the weights are: 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91.”
Now that we know this we can start to visualize the way in which we can obtain the weight of the single bales of hay.After many frustrating tries I realized that the only way to get the correct answer was to use the system below to fill in the blanks.
A+B=80
A+C=82
A+D=83
A+E=84
B+C=85
B+D=86
B+E=87
C+D=88
C+E=90
D+E=91
After discovering this system I plugged in various numbers such as 31, 42, 49, and even 50 the reason that I did this is because our teacher advised us to start in the thirties and not the twenties or lower. I kept trying different possibilities until I got a solution, and in this problems case the solution is:
39+41=80
39+43=82
39+44=83
39+45=84
41+44=85
41+45=86
41+46=87
43+45=88
43+47=90
44+47=91
The numbers are not as orderly as I wished for them to be. It is odd because the system I used to formulate the right answer yields incorrect numbers when I add them all together. Maybe this stroke of luck just saved me three points towards this assignment.
This is the correct way to solve this problem because each two weights add to get the initial group weight. For example,
39+41=80
So on and so forth this goes until we reach:
44+47=91
At this point all of our answers have been solved and we can come to a conclusion for this problem. As far as I know there are no other ways to solve this problem and if there where it seems like they would be harder than the simple adding used in my method of answering this problem. All in all, this is the process I followed to reach my answer to the, “Bales and Bales of Hay” problem.
An extension to this problem could be making the number of bales continue and deviate from the original pattern of solving for the weight. For example you would have to convert bales measured in kilograms to pounds and those measured in pounds to kilograms. A challenging addition to this problem would be to combine decimals and fractions with the already difficult problem. In Conclusion, these are two ways in which the problem could become more challenging and interesting.
In my opinion this problem was a good way to spend our class time, it was challenging enough to leave us stumped for a few days but still easy enough to come up with a clean cut answer to. I thought this problem was relevant and important in bettering our problem solving skills.I wouldn’t do anything to change this problem or others like it and I enjoyed working on it in class. In conclusion, this is my personal reflection of POW 5, “Bales and Bales of Hay.”
Mini POW
December 10th
In this mini Pow students were tasked with solving a data table depicting the number of ice cream scoops on the cone and the exact number of ways that they can be rearranged. For me this problem was possibly the most difficult we have been tasked with so far this year. This problem tricks you by starting out with very easy translation between the numbers of scoops and the way to arrange them. Soon afterwards the difficulty sets in and the numbers grow immensely, at this point I was confused and gave up on the problem for the rest of the day. The next day I worked with lauren to figure out the easiest way to answer numbers five through ten; after the problem was put on the board I realized that the answer could be reached by multiplying the ways to arrange by the number of scoops on the next part:
example:
1x2
2x3
6x4
24x5
ect.
1
1
2
2
3
6
4
24
5
120
6
720
7
5040
8
40320
9
362880
10
3628800
After finding this formula out the rest of the problem basically solved itself.
If we wanted to continue this problem to 50 or even 100 I would continue using this formula to find the answer out. A more in depth look at this formula reveals that it’s use could be prolonged to infinite problems. Just multiply 3628800 by 11 then 39916800 by twelve so on and so forth.
I enjoyed this problem thoroughly and thought it was very challenging and could even be applied to real world. It could have been changed into a different type of story instead of ice cream, maybe shopping for food or buying houses, but those are just minor gripes about this great problem.
In conclusion, this is an example of how you can apply a formula to solve a table problem and could even be applied in the real world.
In this mini Pow students were tasked with solving a data table depicting the number of ice cream scoops on the cone and the exact number of ways that they can be rearranged. For me this problem was possibly the most difficult we have been tasked with so far this year. This problem tricks you by starting out with very easy translation between the numbers of scoops and the way to arrange them. Soon afterwards the difficulty sets in and the numbers grow immensely, at this point I was confused and gave up on the problem for the rest of the day. The next day I worked with lauren to figure out the easiest way to answer numbers five through ten; after the problem was put on the board I realized that the answer could be reached by multiplying the ways to arrange by the number of scoops on the next part:
example:
1x2
2x3
6x4
24x5
ect.
1
1
2
2
3
6
4
24
5
120
6
720
7
5040
8
40320
9
362880
10
3628800
After finding this formula out the rest of the problem basically solved itself.
If we wanted to continue this problem to 50 or even 100 I would continue using this formula to find the answer out. A more in depth look at this formula reveals that it’s use could be prolonged to infinite problems. Just multiply 3628800 by 11 then 39916800 by twelve so on and so forth.
I enjoyed this problem thoroughly and thought it was very challenging and could even be applied to real world. It could have been changed into a different type of story instead of ice cream, maybe shopping for food or buying houses, but those are just minor gripes about this great problem.
In conclusion, this is an example of how you can apply a formula to solve a table problem and could even be applied in the real world.
Cookies Cover Letter:
In the cookies unit we were tasked with creating graphs of feasible regions given certain constraints. In the beginning we started by being presented with the,”cookie problem” which uses many constraint to find the feasible region solving how many cookies the, “ woo’s” could make. Using this core problem we were able to practice and grow using this new type of graphing to solve many problems using all different kinds of constraints. During this unit I was introduced to many key ideas that I will continue to use throughout high school. In conclusion, this was a fun unit to participate in and I enjoyed the problems and am thankful for the help I received from Lauren.
Solution to Problem:
You will need to make 75 plain cookies and 50 iced cookies. You will use 110 pounds of dough to complete this process, you will need 20 pounds of icing and all 15 hours will be used; in addition you will make 125 dozen cookies.Your maximum profit is 212.5 dollars it is so because I graphed the feasible cookie region and I found out if you try for any more and you would push past your available constraints.
Solution to Problem:
You will need to make 75 plain cookies and 50 iced cookies. You will use 110 pounds of dough to complete this process, you will need 20 pounds of icing and all 15 hours will be used; in addition you will make 125 dozen cookies.Your maximum profit is 212.5 dollars it is so because I graphed the feasible cookie region and I found out if you try for any more and you would push past your available constraints.
Pow explanation:
This semester we have done many problem of the week (pow) questions as demonstrated above by the ice cream pow problem, out of all of these Pow's my favorite was the fifth, " Bales and Bales of Hay"
Here is an example of how I solved it directly from the write up itself:
A+B=80
A+C=82
A+D=83
A+E=84
B+C=85
B+D=86
B+E=87
C+D=88
C+E=90
D+E=91
This is the correct way to solve this problem because each two weights add to get the initial group weight. For example,
39+41=80
So on and so forth this goes until we reach:
44+47=91
At this point all of our answers have been solved and we can come to a conclusion for this problem. As far as I know there are no other ways to solve this problem and if there where it seems like they would be harder than the simple adding used in my method of answering this problem. All in all, this is the process I followed to reach my answer to the, “Bales and Bales of Hay” problem.
Here is an example of how I solved it directly from the write up itself:
A+B=80
A+C=82
A+D=83
A+E=84
B+C=85
B+D=86
B+E=87
C+D=88
C+E=90
D+E=91
This is the correct way to solve this problem because each two weights add to get the initial group weight. For example,
39+41=80
So on and so forth this goes until we reach:
44+47=91
At this point all of our answers have been solved and we can come to a conclusion for this problem. As far as I know there are no other ways to solve this problem and if there where it seems like they would be harder than the simple adding used in my method of answering this problem. All in all, this is the process I followed to reach my answer to the, “Bales and Bales of Hay” problem.