# Twelve weights

A classic weighing problem, much harder than the similar Nine weights puzzle.

## Puzzle

Each year at the Fairytale Fair, a contest is held to see which goose can lay a golden egg. Unfortunately, almost everyone who enters just takes a standard silver egg and paints it gold. This year, there are exactly one dozen eggs submitted. The wizard in charge of the contest has a vision, in which it is revealed that all but one egg is a fake. Each fake egg weighs exactly the same, but the real egg weighs either more or less than each of the fake eggs. This being a Fairytale Fair, the only measuring instruments are standard two side balance scales. How can you determine which egg is the real one (and whether it is heavier or lighter) using only the balance scale, and using it only three times?

## Help

**Hint**

**Solution**

- In the first case, we know that the golden egg is among the 4 eggs not weighed and all 8 other eggs are fakes. Take three of the potentially golden eggs and weigh them against three of the fake eggs. Again, the scales could balance or tip. If the scales balance, that means the 4th potentially golden egg is in fact the golden egg. The third weighing can determine whether it is heavy or light. If the scales tip, then we know the golden egg is among the three on the scale, and the direction the scale tips indicates whether it is light or heavy. We can determine which of those 3 eggs is the golden egg using a single scale by placing 1 vs 1 with 1 off. If the scales balance, then the 1 off is golden. If the scales tip, then depending on whether the golden egg was known to be heavy or light, the golden egg is either on the top of bottom side.
- In the second case, we know that the 4 eggs not weighed are fakes, and that the golden egg is either heavy and among the heavy side, or light and among the light side. So we have 4 potentially heavy eggs, and 4 potentially light eggs. Take two potentially heavy eggs and one potentially light egg, and weigh them against a potentially heavy, a potentially light, and a fake egg. Now there are three things that can happen.
- Subcase 1: the scales balance. This can only happen if the golden egg is among the three eggs not on the second scales. There are two potentially light, and one potentially heavy eggs. Weigh the two potentially light eggs against each other. If the scale tips, the top side is the (light) golden egg. If the scale is balanced, the third egg is the (heavy) golden egg.
- Subcase 2: the side with the one fake egg is heavier. This means that either the potentially heavy egg on that side is the (heavy) golden egg, or the potentially light egg on the other side is the (light) golden egg. By weighing the potentially heavy egg against a known fake, we can determine which is the case.
- Subcase 3: the side with the one fake egg is lighter. This means that either the potentially light egg on that side is the (light) golden egg, or one of the two potentially heavy eggs on the other side is the (heavy) golden egg. Weigh the two potentially heavy eggs against each other. If the scales balance, then the other egg is the (light) golden egg. If the scales tip, the bottom side is the (heavy) golden egg.

## See also

Nine weights - an easier version (you know that the different weight is heavier).

Six weights - Find the three heavier weights among three sets of two weights each.