# Imp Pow 15: 12 Bags of Gold

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Imp Pow 15: 12 Bags of Gold
Problem Statement

There are twelve items numbered 1 through 12. All of the values or "weights" are the same except one item whose value is either greater than or less that the other 11 by an unknown amount.

One can compare the sum of the values of a number of items in a set with the sum of the values of items in a disjoint set to see which one is greater. This comparison is also called "weighing."

Find the least number of ways to determine which item has a greater or lesser value.

Process

From the previous POW it was concluded that it would take 5 weighings as described by the equation for an unknown item value:

However, this is an overestimate as there is a way to determine the lighter of nine items with only two weighings. This was overlooked in the last POW (Eight Bags of Gold). Nine items can be weighed by dividing into three sets of size three. Comparing two sets together would determine which set contained the lighter item. With the last weighing, one can determine the lightest item.

This raises the limit of weighings meaning that it would take a mere three weighings to single out the different item out of a set of 27.

Anyway, determining the item which is either lighter or heavier out of 12 is claimed to need only three weighings. It can be done in four because the largest power of three that's within 12 is 33, hence taking three to determine which item is the faulty one. Because the item isn't known to be greater or lesser in value than the other 11, it takes one more. This sets a limit at 4 weighings, at most.

The previous method described goes by a method where the original set of 12 is divided into four sets of three. This is the most efficient method if the sets aren't remixed when measuring. This is also the most efficient if only an imbalance is noticed, and not the direction of the imbalance. To make the method of singling out the unknown valued item out of 12

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