# Imp 1 Pow 14: 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...

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