IMP POW 1: The Broken Eggs
Problem Statement:
A farmer’s cart hits a pothole, causing all her eggs to fall out and break. Luckily, she is unhurt. To cover the cost of the eggs, her insurance agent needs to know how many she had. She can’t remember the number, but can remember some problems she had when packing the eggs. When she put the eggs in groups of two to six eggs, there was always one left over. However, in groups of seven, there were none left over. From what she knows, how can she figure out how many eggs she had?

Process:
First, I thought the answer would be forty-nine. Then, I realized my mistake and tried to think of different ways to do it. I decided to make a chart showing the remainders of the numbers two to six into multiples of seven. I then started to find a pattern. I noticed that the number two has a remainder of one every other multiple of seven, the number three has a remainder of one every two multiples, the number four has a remainder of one every three multiples, and so on. I marked a dot every time there was a remainder of one. I knew that when I had dots marked from two through seven, that would be the answer. So I set out on the long journey of calculating when there was a remainder of one until I reached a number that was all filled up.

Solution:
Using my process, I found that the number of eggs the farmer had was three-hundred one. I know this because it was the first number to fit my boundaries. This number, when put in groups of two to six, will have a remainder of one, but when put in groups of seven, will magically have no remainder. I can assume that there would be another possible larger number that would work, but unfortunately I was too lazy efficient to calculate it.

Evaluation:
Trying to figure out how to solve this problem was very hard. However, once I knew how, it was easier. I think this problem definitely benefited me. I learned that no matter how hard a problem looks. You can always almost solve it. I...

... 09-17-10
Period 5
The BrokenEggsPOW #11. Problem Statement:
A farmer has some eggs in a cart, and is going to market them. She accidently breaks every egg. She doesn’t remember how many she had, but she remembers some things. She knows that when she put them in groups of 2, 3, 4, 5 and 6, there was oneegg left over. When she put them in groups of 13 no eggs were left over. You need to find out how many eggs there are in total.
2. Process:
I first thought about a common multiple of 2, 3, 4, 5, and 6. The least common multiple of those numbers is 60. I know the amount of eggs the farmer has must be a multiple of 60 plus 1. It has to be a multiple of 60 plus 1, because any multiple of 60 plus 1 divided by 2, 3, 4, 5, and 6 will result with a remainder of 1. Then I wrote some multiples of 60 and added 1.
60 | + | 1 | = | 61 |
120 | + | 1 | = | 121 |
180 | + | 1 | = | 181 |
240 | + | 1 | = | 241 |
300 | + | 1 | = | 301 |
360 | + | 1 | = | 361 |
420 | + | 1 | = | 421 |
480 | + | 1 | = | 481 |
540 | + | 1 | = | 541 |
600 | +...

...1. To find my conclusions I had to think about each part of the problem. When you know that one thing means you go on to the next part. When you figure out what that means you have to see how the two statements are related. If they are related then you can deduce a conclusion that makes sense.
2. Here are my conclusions for the 6 problems on page 7.
1. a. No medicine is nice
b. Senna is a medicine
Here I deduced that Senna is not a nice medicine. I think this because the first statement says that “no medicine is nice.” That tells me that all medicines are not nice. The second statement says “Senna is a medicine”. That statement is straight forward. When you put them together you can decide that Senna is a medicine and medicines are not nice. So Senna is not nice.
2. a. All shillings are round
b. These coins are round
Here I decided that no now conclusions can be drawn. The first statement says “All shillings are round.” That statement is clear. The second statement says “These coins are round.” This tells you the coin they have are round. When you put these statements together you can see some flaws. They say these coins but you don’t know if any of these coins are shillings. They can be other coins that are round. So you cannot deduce anything.
These coins are
3. a. Some pigs are wild
b. All pigs are fat
Here I decided that there are no conclusions that can be made. The first...

...“A Sticky Gum Problem” POW 4
Problem statement:
The next scenario is very similar. In this one, Ms. Hernandez passed a different gumball machine the next day with three different colors Once again her twins each want a gumball of the same color, and each gumball is still one cent. What is the most amount of money that Ms. Hernandez would have to spend in order to get each of her daughters the same color gumball?
In the last scenario, Mr. Hodges and his triplets pass the same gumball machine that Ms. Hernandez and her twins passed in scenario two. This time, each of Mr. Hodges children wants the same color gumball out of the three-color gumball machine. What is the most amount of money that Mr. Hodges would have to spend on his triplets in order to get them each the same color gumballs?
Process:
In the process of solving the first question I drew up different color gumballs.
These are both the different colored gumballs. If this was the outcome after spending two pennies on two gumballs, then the next gumball would have to be one of the previous color gumballs that already came out of the machine.
For the next scenario it was a little trickier that the first problem. Since there were still only two children involved and there were three colored gumballs it wasn’t too hard. Once again, I drew up the three different colored gumballs in the gumball machine. The gumballs were red, white, and...

...Mega POW
A very wealthy king has 8 bags of gold, which he trusts to some of his caretakers. All the bags have equal weight and contain the same amount of gold, all the gold in the kingdom. Although, the king heard a story that a woman received a gold coin. The king knew it had to be his gold so he wanted to find the lightest bag in the 3 weighing, but the mathematician thought it could be done in less, so I need to find out the least amount of weighing it takes to find the lightest bag. Also, the king used a pan balance for all of his weighing.
I started by weighing 4 bags on each side of the scale to see which side was lighter. Then from those results I thought to weigh the 4 bags that were on the lighter side by 2 and 2. After this you would find one side weighing less than another. Then you would take those results and weigh the 2 remaining bags and the lightest bag would be the bag that was taken from. However, the mathematician said it could be done in less than three steps. So throwing the answer I had just gotten to the side, I started new. This time I started with 3 bags on each side knowing that if two sides were equal than the bag with the missing gold would be one of the bags not weighed the first time. Then you would have to weigh the two remaining bags and whichever one was lighter than the other would be the bag with less gold. But, if the 3 bags from the beginning weighed different then you would weigh 2 bags of the 3 and if they are equal...

...probability of success or the one most likely to help me win.
Strategy # 1
a. Always choose the same thing the card says. So if it is an O choose O, if it is an X choose X.
b. 30 trials
1. yes 6.no 11.no 16.yes 21.yes 26. yes
2. yes 7.yes 12.yes 17.yes 22.no 27. yes
3. yes 8.no 13.yes 18.no 23.yes 28. yes
4. no 9.no 14.yes 19.yes 24.no 29. yes
5.no 10.yes 15.yes 20.no 25.no 30. Yes
P (right) – 18/30 or 6/10 or .6
Success rate = .6
c. (XX)-works-1/6
(XX)-works-1/6
(OO)-works-1/6
(OO)-works-1/6
(OX)-doesn’t work
(XO)-doesn’t work total successful – 4/6 or 2/3 or .66
Success rate = .66
Strategy #2
a. Always choose O no matter what.
b. 30 trials
1. yes 6.no 11.yes 16.yes 21.no 26.yes
2. yes 7.no 12.yes 17.yes 22.no 27.no
3. no 8.yes 13.no 18.yes 23.no 28.yes
4. no 9.no 14.yes 19.yes 24.yes 29.yes
5.no 10.yes 15.yes 20.yes 25.no 30.no
P (right) – 17/30 or .56
Success rate = .56
c. (XX)-doesn’t work
(XX)-doesn’t work
(OX)-doesn’t work
(XO)-works-1/6
(OO)-works-1/6
(OO)-works-1/6 total successful – 3/6 or .5
Success rate = .5
Strategy # 3
a. Always choose the opposite of the card...

...POW (BrokenEggs)
A farmer is taking her eggs to the market when her cart tips over shattering all of the eggs. She goes to an insurance agent unsure of how many there actually were. She needs to know this to tell the agent though. However, she does know that when she put the eggs in groups of two, there would be one left over. This also seemed to be true for groups of three, four, five, and six. But when she put them in groups of seven, there were an equal number of eggs in each group with none left over.
At first glance, some people might assume the answer is 49 because from simply looking at the problem you would say “Oh, 7 goes evenly into 49, with nothing left over” and from quick thinking you would assume the other numbers would too. But when you try and do the math, you realize that all of the numbers need to go into 48 with 1 left over, and most do, but 5 does not go evenly go into 48 with one left over. So the answer could not possibly be 49.
I began this problem not really sure how to start. I knew that the answer would have to be a multiple of seven so I went from there. Starting from 49 (because I thought it wouldn’t be 49 or lower), I tried each multiple of seven until I got to 140. This is when I started to rethink my strategy, I knew there had to be a more logical way of looking at this. I talked to my friend Michelle who made it clear...

...Name: Modella Studente POW # 1BrokenEggs
Problem Statement: How many eggs were broken? (And is there more than one answer?)
Process: Given that she lined them up by twos and one was left over, by threes and one was left over, by fours and one was left over, by fives and one was left over, by sixes and one was left over and by sevens and it came out evenly, we figure the number had to be a multiple of seven and end in a one or a six. (Anything divisible by five with one left over has to end in a one or six: 6, 11, 16, 21, etc. We figure that out when we eliminated 49 as an answer.)
So we started with 21 (the first multiple of seven that ends with a one or a six) and found you couldn't divide by three and have one left over. We went up by sevens until we found 56 (but when we divided by two we didn't have one left over) and 91 (but when we divided by 4 we didn't have one left over) and 136 (which didn't work for two) until we figure out we could go up by 35 to reach the next multiple of seven that ends with a one or a six.
Going up by 35's (161, 206, 231, 276, …) finally led us to 721. An answer that worked! Everyone in the group double-checked it and we found an answer. Now is it the only answer. We realized that the answer had to be odd (because any number divided by two with one left over is odd) so we could go up by 70's.
So we started a new list (791, 861, 931,...

...Pow1
10/6/10
Pow
A farmer is taking her eggs to the market in a cart, but she hits a Pothole, which knocks over all the containers of eggs. When she put the eggs in groups of two, three, four, five, and six there was one egg left over, but when she put them in groups of seven they ended up in complete groups with no eggs left over. Now she needs to know how many eggs she had and is there more than one possibility.
The first thing I did was to read the pow aging on my own. I out when she put her eggs in groups of two there is one left over. The number cannot be a multiple of two. Also three four five and six can’t be a multiple of this number. If there were no eggs left over when put into groups of seven there must have been a multiple of 7 eggs. Now need to find multiples of seven.
7,14,21,28,35,42,49,56,63,70,77, 84,91,98, 105,112,119,126,133,140,147 ,154,161,168,175,1 82 ,189,196 ,203 ,210, 217, 224, 231 ,238, 245, 252 ,259, 266, 273, 280,287,294,301
Then you cross out all the numbers that are divisible by 2,3,4,5, and 6. So I got161 and 301 as the numbers that cannot be multiples 2, 3,4,5,6.
| 3 * 4 * 7 = 8449 + 84 = 133. No good. 133 is not good because it is not a multiple of 7133 + 84 = 217. No good. 217 because it is not a multiple of 7217 + 84 =...