# Imp 1 Pow 14: Mega Pow

By powowgirl12
Mar 26, 2005
972 Words

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 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 of bags with different objects in them. Just changing it so there is variation.

This POW taught me that you couldn't just assume an answer and think its right you need to try different strategies in order to find the true minimum of a problem. Also, this POW flexed the skill of organized thinking and planning because you have to keep all the information from all the ways tried to understand why 3 is not the minimum number.

We return again to the king only this time with a much more difficult task. The king has found another untrustworthy caretaker who has been making counterfeit gold, they found her but still have some issues. The woman only told the king that it was one of his 12 bags of gold since at this time he uncovered more gold and put them into 12 bags instead of 8. They only know that the one bag would weigh different than the other bags, they need to find whether it would weigh more or less. Now the king wants this done in only 2 weighing like the 8 bags but the court mathematician said the least that could be done is 3. I have to find in as least amount of weighs which bag is counterfeit.

To start off I used a diagram to get into my thought process by drawing the scale with 6 bags on each side. Then you would see one side that would weigh less and one that would weigh more. I started with the side with the bags weighing less. What I did was split the bags in half again. If both sides weighed equal then you would know that none of those bags could possibly have the counterfeit money and you would move on to the heavier bags from the first weighing and split those in half, with 3 on each side. They can't weigh the same if the other 6 bags did, so what you would do is take the lighter side of the bags and weigh them with the 3 bags that all weighed the same from the other weighing and if they are still heavier then you know the one of the bags is the counterfeit and you would just have to weigh once or twice more to find which one. But once again if the bags all weigh the same then you would move onto the other 3 bags that weighed more and you would know that one of them is the counterfeit and you would way for the third time and find which it is. Although if you started with the heavier side in the first weighing you would do a similar process with what I did with the lighter bags and would still end up finding the counterfeit bag. If in the second weighing process if the bags didn't weigh equally you would do the same process you did on the third weighing.

This process should work every time you have a similar situation because it involves 3 to 4 weighs and will always end up with finding the counterfeit bag even if you don't know if it weighs more or less than the other bags. This Mega POW helped me better understand shorter weighs of finding things and made me work on my strategy skills. I'm building more brain cells if I could grade it, I would give it an A-.

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 of bags with different objects in them. Just changing it so there is variation.

This POW taught me that you couldn't just assume an answer and think its right you need to try different strategies in order to find the true minimum of a problem. Also, this POW flexed the skill of organized thinking and planning because you have to keep all the information from all the ways tried to understand why 3 is not the minimum number.

We return again to the king only this time with a much more difficult task. The king has found another untrustworthy caretaker who has been making counterfeit gold, they found her but still have some issues. The woman only told the king that it was one of his 12 bags of gold since at this time he uncovered more gold and put them into 12 bags instead of 8. They only know that the one bag would weigh different than the other bags, they need to find whether it would weigh more or less. Now the king wants this done in only 2 weighing like the 8 bags but the court mathematician said the least that could be done is 3. I have to find in as least amount of weighs which bag is counterfeit.

To start off I used a diagram to get into my thought process by drawing the scale with 6 bags on each side. Then you would see one side that would weigh less and one that would weigh more. I started with the side with the bags weighing less. What I did was split the bags in half again. If both sides weighed equal then you would know that none of those bags could possibly have the counterfeit money and you would move on to the heavier bags from the first weighing and split those in half, with 3 on each side. They can't weigh the same if the other 6 bags did, so what you would do is take the lighter side of the bags and weigh them with the 3 bags that all weighed the same from the other weighing and if they are still heavier then you know the one of the bags is the counterfeit and you would just have to weigh once or twice more to find which one. But once again if the bags all weigh the same then you would move onto the other 3 bags that weighed more and you would know that one of them is the counterfeit and you would way for the third time and find which it is. Although if you started with the heavier side in the first weighing you would do a similar process with what I did with the lighter bags and would still end up finding the counterfeit bag. If in the second weighing process if the bags didn't weigh equally you would do the same process you did on the third weighing.

This process should work every time you have a similar situation because it involves 3 to 4 weighs and will always end up with finding the counterfeit bag even if you don't know if it weighs more or less than the other bags. This Mega POW helped me better understand shorter weighs of finding things and made me work on my strategy skills. I'm building more brain cells if I could grade it, I would give it an A-.