Pow 14- King's Gold

POW 14 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. 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 fewer coins. If they weigh differently, the lighter bag would be the one with the least amount of 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.

Solution:

Weigh only 6 out of the 8 bags, with 3 bags on each side of the scale.

3 Ð'- 3

If the bags are equal, one of the 2 remaining bags of gold is the one missing coins; from there, you could just weigh the two remaining bags of gold and the one that weighs less will be the bag with missing gold.

But if one of the 3 on either side of the scale is missing coins, the scale will be uneven, and unbalanced.

In this case you would take 2 out of the 3 bags of gold and weigh those. If the 2 you weighed are even, then the bag of gold you didn't weigh will be the bag with the missing coins.

1 Ð'- 1

1

If the bag with the missing coins IS one of the ones you weighed, then obviously the bag with the lesser weight is the one missing coins.

Extensions: (Ð'...Don't be surprised if this sounds familiar- Russ helped me with this again. ^^)

After this past predicament with the King and his gold, the king has decided to divide his gold further- into 12 bags instead of 8, so that the caretakers will have less to be tempted by. But despite this change, the king has found yet another untrustworthy caretaker for his gold- this time, the caretaker spending his money is trying to fool him as well. After spending his money, they are replacing his