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pow 5
Process: Starting this POW I arranged the combinations added weights of the hay bales.next I started to look for an equation that would give me each weight. I tried looking for a way to take the combinations with the same hay bale. After a while of looking at these possibilities and realizing that it was probably possible to to figure it out this way I decided there must have been a different way to figure it out. so I then started looking for at the pattern in the added bale weights, in that they skipped two up the between the first and second combination but after continued up by one each time. Since there could be no two of the same I figured that bale 1 would have to have the smallest weight and that the rest had to go up by one each time, and I came up with what is in the second graph below.
b1+b2
80
b1+b3
82
b1+b4
83
b1+b5
84
b2+b3
85
b2+b4
86
b2+b5
87
B1
39
B2
41
B3
42
B4
43
B5
44
How many squares?
Process: To start this process I made an 8x8 square on a piece of graph paper. Second, I counted how many 7x7 squares could fit into the larger square. Then I did the same with 6x6 5x5 4x4 and so on.
8=1
7=4
6=9
5=16
4=25
3=36
2=49
1=64
To find an equation for any size square I tried looking for a pattern in the number of smaller squares that fit into the large one. after I didn’t find anything trying that I looked at how those same numbers corresponded with the number of individual squares at the smallest level. Finally I looked at how many of the different squares that fit going one way the 2x2 squares fit 7 in one row. the total number of 2x2 squares that fit into the 8x8 square is 49 which I noticed is just 7 times 7. I tried this with the other size squares and found the same was true for them.
Solution: I found that if you count the number of whatever size square fits into a row, and square that you will have the total number of how many total squares of that size will fit. For example 5 4x4 squares fit in one row take that 5 and times it by 5 and you get 25 which is the total number of squares that fit into the 8x8 square.
answer: 81
I did enjoy working on this problem because it had some good, practical problem solving.
it was not too hard of a problem to solve, but it did take more time than most problems. To change this problem I think adding a second size square would further prove the ability to find how many squares without counting.
8=1
7=4
6=9
5=16
4=25
3=36
2=49
1=64
To find an equation for any size square I tried looking for a pattern in the number of smaller squares that fit into the large one. after I didn’t find anything trying that I looked at how those same numbers corresponded with the number of individual squares at the smallest level. Finally I looked at how many of the different squares that fit going one way the 2x2 squares fit 7 in one row. the total number of 2x2 squares that fit into the 8x8 square is 49 which I noticed is just 7 times 7. I tried this with the other size squares and found the same was true for them.
Solution: I found that if you count the number of whatever size square fits into a row, and square that you will have the total number of how many total squares of that size will fit. For example 5 4x4 squares fit in one row take that 5 and times it by 5 and you get 25 which is the total number of squares that fit into the 8x8 square.
answer: 81
I did enjoy working on this problem because it had some good, practical problem solving.
it was not too hard of a problem to solve, but it did take more time than most problems. To change this problem I think adding a second size square would further prove the ability to find how many squares without counting.
the phone tree
Process: Starting this POW I tried only 2 methods to just think of how many of Charlotte’s friends could be called by 9:00 if each of them called 2 more friends at a rate of 3 minutes a friend. First I tried to think of a way to use multiplication, but without success. Next asked my dad what he thought of the problem and he said it would be a good idea to draw it. So I tried to use a tree graph to be able to physically show how a line of people calling people would work. when this started working I saw that every 2 minutes the number of people called doubled; so I started doing that, and got my solution below.
Solution: 1,048,576
Evaluation: I found this to be a very worthwhile because it helped me see how exponential growth can work so quickly. I enjoyed working on this POW because I enjoy working on problems without a specific way to solve, but that instead are simply something to figure out. Another thing I enjoyed about this POW is how surprising the answer I got was.
Solution: 1,048,576
Evaluation: I found this to be a very worthwhile because it helped me see how exponential growth can work so quickly. I enjoyed working on this POW because I enjoy working on problems without a specific way to solve, but that instead are simply something to figure out. Another thing I enjoyed about this POW is how surprising the answer I got was.
POW Free thinking football.
Logan Shackles
period 5
What I did to solve the initial problem was, first start at 3 and find all the numbers that are not possible to make with either 5 or 3 by looking at each number above 3 and checking to see if it was possible to make. The highest number I found was 7. On a second problem I used 6 and 4 and that did not have a highest impossible score because they are both even numbers. I repeated this two more times, once with my group with 5 and 9 which had a highest impossib-le number of 31, and once by myself with 4 and 11 which had a highest impossible number of 79. I considered this educationally worthwhile in that I learned not all the POW’s are gonna have some formula that makes finding the answer really fast. this was not a particularly hard POW ut it did take a lot of time get the answer for each set of numbers.
period 5
What I did to solve the initial problem was, first start at 3 and find all the numbers that are not possible to make with either 5 or 3 by looking at each number above 3 and checking to see if it was possible to make. The highest number I found was 7. On a second problem I used 6 and 4 and that did not have a highest impossible score because they are both even numbers. I repeated this two more times, once with my group with 5 and 9 which had a highest impossib-le number of 31, and once by myself with 4 and 11 which had a highest impossible number of 79. I considered this educationally worthwhile in that I learned not all the POW’s are gonna have some formula that makes finding the answer really fast. this was not a particularly hard POW ut it did take a lot of time get the answer for each set of numbers.