# Pow 14- King's Gold

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...

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