Rats POW
In this problem, two rats, one male and one female, find themselves on a deserted island. We were tasked with finding the amount of rats on the island after one year, given these assumptions.
-The female has a litter of six babies on January 1. She keeps making liters of rats every 40 days until she dies.
-Every female rat on the island starts having babies 120 days after they are born, and continue having them every 40 days from then.
-Every litter has 3 males and 3 females
-No rats die in the first year.
To start this problem, I first broke up the year into nine 40-day chunks by dividing 360 by 40 and rounding down. I then wrote out the amount of rats the first day, 8 with one reproducing. I then multiplied the one reproducing rat by six and adding the result to the original 8 to find the amount of rats after 40 days. For the next 40 days I did the same, giving me 20 rats in 80 days. For the next 40 days, I found there were 4 rats reproducing now, because it had been 120 days since the first 3 rats were born. I took the 4, multiplied by 6 and added it to the 20 giving me 44 rats. The next chunk, the rats from the second chunk started reproducing, making 7 litters for a total of 86 rats. Next, I took the amount of rats from 2 chunks back, divided by 2 because only half were female, and multiplied by 6 for a total of 146 rats. I did the same for the rest of the chunks, so there was 278 at 240 days, 536 at 280 days, 974 at 320 days, and 1,808 at 360 days.
A habit of a mathematician I used extensively during this POW was staying organized because the problem has a lot going on and it was tough keeping everything sorted out and making sense while I was working on it. I am confident in my answer because at the beginning of the problem we were given the final number and my work showed the same number in the end.
-The female has a litter of six babies on January 1. She keeps making liters of rats every 40 days until she dies.
-Every female rat on the island starts having babies 120 days after they are born, and continue having them every 40 days from then.
-Every litter has 3 males and 3 females
-No rats die in the first year.
To start this problem, I first broke up the year into nine 40-day chunks by dividing 360 by 40 and rounding down. I then wrote out the amount of rats the first day, 8 with one reproducing. I then multiplied the one reproducing rat by six and adding the result to the original 8 to find the amount of rats after 40 days. For the next 40 days I did the same, giving me 20 rats in 80 days. For the next 40 days, I found there were 4 rats reproducing now, because it had been 120 days since the first 3 rats were born. I took the 4, multiplied by 6 and added it to the 20 giving me 44 rats. The next chunk, the rats from the second chunk started reproducing, making 7 litters for a total of 86 rats. Next, I took the amount of rats from 2 chunks back, divided by 2 because only half were female, and multiplied by 6 for a total of 146 rats. I did the same for the rest of the chunks, so there was 278 at 240 days, 536 at 280 days, 974 at 320 days, and 1,808 at 360 days.
A habit of a mathematician I used extensively during this POW was staying organized because the problem has a lot going on and it was tough keeping everything sorted out and making sense while I was working on it. I am confident in my answer because at the beginning of the problem we were given the final number and my work showed the same number in the end.
Patches POW
In this problem, we were tasked with finding the maximum amount of rectangles of a certain size we could cut from a larger piece of a certain size. In problem 1, we had to find the amount of 3 by 5 pieces we could make from a 17 by 22 piece. We found that we could make 23 pieces from it. In problem 2 we had to find 9 by 10 pieces from the 17 by 22 piece. We found we could only make 2 pieces. we also had to find 5 by 12 pieces and 10 by 12 pieces, which allowed 5 and 2 pieces, respectively. In problem 3, we had to find 3 by 5 patches from a 4 by 18 piece and an 8 by 9 piece, which showed three and five pieces. We then experimented with various patch sizes and piece sizes. For an 11 by 3 rectangle, we could find sixteen 2 by 1 pieces. For a 20 by 20 square, we found twenty-six 3 by 5 pieces. In general, we found that we couldnt get the maximum number of pieces by lining all the patches up one by one, but had to have a certain amount of them in different orientations to each other. During the course of this problem, our group demonstrated collaboration and visualization by asking each other for our results to make sure we got the maximum number of patches possible, and by drawing diagrams to further understand the shapes and orientations of the various patches.
POW #3
Problem Statement: In this problem, we had to find the equation for finding the area of a shape using the pegs on the outside and inside of a shape.
Process: There were three parts to this problem, finding the formula for the area of a shape with no interior pegs, finding the formula for area of shape with only 4 exterior pegs using the interior pegs, and finding the formula for the area of a shape with any exterior or interior pegs.
I started by looking at examples of shaped with no interior pegs and comparing exterior pegs to area. Using my data, I found the formula,
(P/2)-1. P being the number of perimeter pegs.
For the second part, I used the same method, comparing areas to numbers of interior pegs, and found the formula,
I+1. I being the number of interior pegs.
I then combined the two equations to find the formula for finding area of any shape on a geoboard, and tested it on many different shapes, to make sure the formula worked
Solution: Area= I+(E/2)-1
I being interior pegs and E being exterior pegs
Habits: Look For Patterns: I used this habit to find patterns in the data comparing pegs to area, which I used in creating my formulas.
Double Check: I used this habit when I found my different equations, to test my formulas on different shapes, in case they were incorrect
Process: There were three parts to this problem, finding the formula for the area of a shape with no interior pegs, finding the formula for area of shape with only 4 exterior pegs using the interior pegs, and finding the formula for the area of a shape with any exterior or interior pegs.
I started by looking at examples of shaped with no interior pegs and comparing exterior pegs to area. Using my data, I found the formula,
(P/2)-1. P being the number of perimeter pegs.
For the second part, I used the same method, comparing areas to numbers of interior pegs, and found the formula,
I+1. I being the number of interior pegs.
I then combined the two equations to find the formula for finding area of any shape on a geoboard, and tested it on many different shapes, to make sure the formula worked
Solution: Area= I+(E/2)-1
I being interior pegs and E being exterior pegs
Habits: Look For Patterns: I used this habit to find patterns in the data comparing pegs to area, which I used in creating my formulas.
Double Check: I used this habit when I found my different equations, to test my formulas on different shapes, in case they were incorrect
POW #2
Problem Statement: In this problem, we were tasked with finding the maximum amount of gumballs needed to be purchased before multiple of the same color are found.
Process: In the first variation of this problem, two gumballs of the same color are needed from a possible two different colors, red and white. Because we are looking for the Maximum amount of gumballs, we are looking for the worst-case-scenario outcome, which would be to get two gumballs of different colors at first, and because there are only two colors, the next gumball has to be a second of a color, so the maximum amount is 3.
In the next variation, three gumballs are needed out of two different colors. Worst-case-scenario, there are two gumballs of each color drawn the first four times, and the fifth time, the gumball has to be one of the two colors, so the maximum amount is 5.
In the third variation, two gumballs of the same color are needed, but out of three colors this time. Worst-case-scenario, one gumball of each color is drawn the first three times. The fourth time, the gumball has to be one of the three colors already drawn, so the maximum amount of gumballs needed is 4.
Finally, the formula for the maximum amount of gumballs is needed. To start off, the variables are decided. For this equation, I am using
X=the amount of gumballs of the same color needed,
Y=the amount of colors possible
and Z= the maximum gumballs needed to be drawn to reach X.
Solution: I started off by looking at the problems already solved, and found that X increases Z more than Y does. I then found that an odd Y makes Z an even #. Using this information, I found the equation,
Z= [(X-1) *Y]+1
After finding this equation, I began testing it with various combinations of gumball colors and gumballs needed, 3 gumballs and 3 colors gives a maximum of 7, 4 colors and 3 gumballs gives a maximum of 9, 4 and 4 give 13, and so on.
Habits: Look for patterns: One of the habits I used when solving this problem was looking for patterns. I used the data from the first few parts of the problem to find my equation, making finding the final answer much easier.
Experiment through conjectures: Another habit I used when solving this problem was experimenting. I experimented with different combinations of gumballs and colors to make sure my formula was correct. This helped me make sure I didn't have a wrong answer.
Process: In the first variation of this problem, two gumballs of the same color are needed from a possible two different colors, red and white. Because we are looking for the Maximum amount of gumballs, we are looking for the worst-case-scenario outcome, which would be to get two gumballs of different colors at first, and because there are only two colors, the next gumball has to be a second of a color, so the maximum amount is 3.
In the next variation, three gumballs are needed out of two different colors. Worst-case-scenario, there are two gumballs of each color drawn the first four times, and the fifth time, the gumball has to be one of the two colors, so the maximum amount is 5.
In the third variation, two gumballs of the same color are needed, but out of three colors this time. Worst-case-scenario, one gumball of each color is drawn the first three times. The fourth time, the gumball has to be one of the three colors already drawn, so the maximum amount of gumballs needed is 4.
Finally, the formula for the maximum amount of gumballs is needed. To start off, the variables are decided. For this equation, I am using
X=the amount of gumballs of the same color needed,
Y=the amount of colors possible
and Z= the maximum gumballs needed to be drawn to reach X.
Solution: I started off by looking at the problems already solved, and found that X increases Z more than Y does. I then found that an odd Y makes Z an even #. Using this information, I found the equation,
Z= [(X-1) *Y]+1
After finding this equation, I began testing it with various combinations of gumball colors and gumballs needed, 3 gumballs and 3 colors gives a maximum of 7, 4 colors and 3 gumballs gives a maximum of 9, 4 and 4 give 13, and so on.
Habits: Look for patterns: One of the habits I used when solving this problem was looking for patterns. I used the data from the first few parts of the problem to find my equation, making finding the final answer much easier.
Experiment through conjectures: Another habit I used when solving this problem was experimenting. I experimented with different combinations of gumballs and colors to make sure my formula was correct. This helped me make sure I didn't have a wrong answer.
POW #1
Problem Statement: In this problem, we were tasked with creating an equation resulting in the numbers from 1 to 25 using only the numbers 1,2,3,and 4 exactly once.
For example, if you were to try to find 10, you would simply put 1+2+3+4 as it uses each number exactly once, and results in the number 10.
Process: First, we decided to go one at a time and just work them out in our head, and collaborated to help each other with problems they were stuck on. This was successful, so I'll give you the solutions that we found.
Solutions:
1) 1(3+2)-4
2) 1+2+3-4
3) 3!-4-1+2
4) 1+2-3+4
5) (1+2)*3-4
6) 3^2-4+1
7) (3*4)/2 +1
8) (4^1/2 *2)(3-1)
9) (4*3)-2-1
10) 1+2+3+4
11) (2*3)+4-1
12) 3^2 +4-1
13) (4*3)+(2-1)
14) 2(3+4)/1
15) (4!)/2 +3/1
16) 3+1^2*4
17) 3(4+1)+2
18) 4^(1/2)(3+2)/1
19) 4(3+2)-1
20) 4(3+2)/1
21) (4+3)(1+2)
22) 4!-3+(1^2)
23) 2(3*4)-1
24) 2(3*4)/1
25) 2(3*4)+1
Habits:
Be Systematic: We went through the problems one at a time, not proceeding until we finished the problem before it. We did this method to make sure that we didn't accidentally skip a number.
Collaborate and Listen: We would ask others for help when we were stuck on a problem, and helped others in turn when they were stuck on a problem. We did this to make sure that we had all the correct problems, but still made sure everyone did their fair share of work.
For example, if you were to try to find 10, you would simply put 1+2+3+4 as it uses each number exactly once, and results in the number 10.
Process: First, we decided to go one at a time and just work them out in our head, and collaborated to help each other with problems they were stuck on. This was successful, so I'll give you the solutions that we found.
Solutions:
1) 1(3+2)-4
2) 1+2+3-4
3) 3!-4-1+2
4) 1+2-3+4
5) (1+2)*3-4
6) 3^2-4+1
7) (3*4)/2 +1
8) (4^1/2 *2)(3-1)
9) (4*3)-2-1
10) 1+2+3+4
11) (2*3)+4-1
12) 3^2 +4-1
13) (4*3)+(2-1)
14) 2(3+4)/1
15) (4!)/2 +3/1
16) 3+1^2*4
17) 3(4+1)+2
18) 4^(1/2)(3+2)/1
19) 4(3+2)-1
20) 4(3+2)/1
21) (4+3)(1+2)
22) 4!-3+(1^2)
23) 2(3*4)-1
24) 2(3*4)/1
25) 2(3*4)+1
Habits:
Be Systematic: We went through the problems one at a time, not proceeding until we finished the problem before it. We did this method to make sure that we didn't accidentally skip a number.
Collaborate and Listen: We would ask others for help when we were stuck on a problem, and helped others in turn when they were stuck on a problem. We did this to make sure that we had all the correct problems, but still made sure everyone did their fair share of work.
Obama Inauguration Problem
Problem Statement: In this problem, we were tasked with finding the number of people at Obama's inauguration in 2008 using only a satellite image of Washington D.C. for reference.
Method: First, we tried to get a scale by looking up the actual distance from the Washington Monument to the Capital Building, and then measuring the distance on the satellite image. We then divided the measured distance by the actual distance, and got a scale of 0.064Km/cm on the map.
Then, we took the average population density of a crowd, 2 people/m^2, and, using the scale, found the area of the crowds.
230,400 m^2
We then multiplied this number by two to give us the number of people.
460,800 People
But, the crowds weren't taking up the entire space, only about 1/3 of it. So, subtracting 2/3 from the 460,800 people, we got our final number,
Solution:
153,600 People
Habits:
Collaborate and listen: We came together and valued the ideas of everyone in the group, we all contributed ideas, and built off of each other equally. We never flat out rejected someone else's ideas and always considered any solutions that came up.
Solve a simpler problem: We broke the problem down into smaller parts and took them on one by one. We found small steps to take so the problem never became too overwhelming for the group to manage. We took the bigger, overall problem and broke it down to find the solution.
Method: First, we tried to get a scale by looking up the actual distance from the Washington Monument to the Capital Building, and then measuring the distance on the satellite image. We then divided the measured distance by the actual distance, and got a scale of 0.064Km/cm on the map.
Then, we took the average population density of a crowd, 2 people/m^2, and, using the scale, found the area of the crowds.
230,400 m^2
We then multiplied this number by two to give us the number of people.
460,800 People
But, the crowds weren't taking up the entire space, only about 1/3 of it. So, subtracting 2/3 from the 460,800 people, we got our final number,
Solution:
153,600 People
Habits:
Collaborate and listen: We came together and valued the ideas of everyone in the group, we all contributed ideas, and built off of each other equally. We never flat out rejected someone else's ideas and always considered any solutions that came up.
Solve a simpler problem: We broke the problem down into smaller parts and took them on one by one. We found small steps to take so the problem never became too overwhelming for the group to manage. We took the bigger, overall problem and broke it down to find the solution.