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 in weight than the 3rd bag is the one with less coins. If they weigh different the lighter bag would be the one with less coins.

The least amount of times of weighing you need to do in order to find the bag with missing gold is 2 because any-other way of problem solving this question would get you 3 or more. I know this because I tried every different possibility.

Another way of practicing this problem solving skill is to have a similar situation but with more bags of gold, maybe even with an odd number...

...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...

...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...

...POW Problem Statement
A. A farmer is going to sell her eggs at the market when along the way she hits a pot hole causing all of her eggs to spill and break. She meets an insurance agent to talk about the incident, and during the conversation he asks, how many eggs did you have? The farmer did not know any exact number, but proceeded to explain to the insurance agent that when she was packing the eggs, she remembered that when she put the eggs in groups of 2-6 she had even groups with 1 left over, but when she put them in groups of 7 she had even groups of 7 with none left over.
B. Why does groups of 2,3,4,5 or 6 results in 1 left over egg, but groups of 7 has an equal amount of eggs with none left over. What # of eggs has equal groups of 2,3,4,5, or 6 with one left over and 7 goes into the number evenly.
C. They think the answer is 49 eggs because 7 goes into 49 eggs evenly with none left over.
D. It cannot be 49 eggs because if it were 49 then 2-6 would need to go into 48 evenly to have a left over egg, but 5 does not go into 48 evenly which is why 49 wouldn’t work
POW Process
A. My initial ideas concerning the task is we need to find a number that has 2,3,4,5 and 6 go into that number evenly with 1 left over and have 7 go into it evenly with none left over. Making a chart with multiples of all those numbers out on paper will help us track a number down and...

...Problem Statement:
Some families didn’t want to travel overland to California so they took ships around Cape Horn at the tip of South America. Say a ship leaves San Francisco for New York the first of every month at noon. At the same time a ship leaves New York for san Francisco. Every ship arrives exactly 6 months after it leaves.
If you were going to San Francisco from New York How many ships from San Francisco would you meet?
I assumed that entering and exiting the harbor does not count as meeting a ship. I also assumed that there was already (only) one ship at see and that it had traveled 3 months when the main ship leaves.
Process:
My first thought were ‘‘ok lets draw a picture” they were shortly followed by “dang i cant draw North and South America”. Then I had an amazing thought “The Americas form a triangle, I can draw a triangle!” Then i though that iI had got it and this was gunning to be E-A-S-Y.
Well first i tried drawing the Americas and failed. I then drew a triangle and labeled months on it. i took little ripped pieces of paper and moved them around the triangle and made a tally every time they met. After i did that i mad a version on the computer which enabled me to move things around and show my thinking better.
I basically thought “ how is she going to understand this” thats when i decided to put the “map” on the computer. I mapped out the paths color coded the ships and the routes across the water. After i had all the arrows in place i...

...Pow14imp1.
conner Douglas
1. Problem statement.
A wealthy king has 8 bags of gold that gives to some of his most trusted friends. All the bags have the same weight and the same amount of coins in the bags is all of the gold in the kingdom. Although, the king herd that a local woman received a gold coin. The king knew that it had to be one of his coins so he wanted to find the lightest bag in 3 weightings. But his court mathematician thought it could be done in less, so I need to find out the least amount of weightings.
2. Process
I started by weighing 4 bags on each side of the simulator scale to see which side was lighter. Then from those results I thought to weigh the 4 bags that were on the lighter side then again I would weigh the 2 lighter ones to find out which one it is. However the mathematician said that I can be done in less than 3 moves, so throwing the answer that I just got to the side, I started new. This time I started with 3 bags on each side knowing that if both sides were equal then the bag with the missing gold will be in one of the two other bags. But I knew that you wouldn’t find out which one is the lightest one on the second time. Then you would take the other 3 bags and way 2 at a time then if they way the same then you know which one is lighter to get the bag with fewer coins.
3.
The least amount of times of weightings you need to do in order to find the...

...POW14 Christopher Manahan
Period 05
February 28, 2006
Problem Statement:
A very wealthy king has 8 bags of gold- all the gold in the kingdom, which he trusts to 8 of his most trustworthy caretakers; one bag to each caretaker. All the bags have equal weight and contain the same amount of gold, totaling all the gold in the kingdom. But one day, the king hears a story that a woman from another kingdom received a gold coin. The king knew it had to be his gold, because he owned all the gold in the kingdom. Someone was spending his gold! So he decided to find the lightest bag of the 8 using a pan scale to weigh the bags of gold.
The King expected that it would take 3 weightings to determine the lightest bag of gold, but the mathematician thinks the lightest can be determined in less. I need to find out the lowest number of times that the King will have to weigh his gold to determine the lightest bag.
Process:
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....

...Emily Shiang
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...