Puzzle: We have two red, two green and two yellow balls. For each color, one ball is heavy and the other is light. All heavy balls weigh the same. All light balls weigh the same. How many weighings on a beam balance are necessary to identify the three heavy balls?
Source: Mathematical Circus (1979, 272 pages) by Martin Gardner.
Solution: Two weighings suffice. Weigh one red and one green ball against one yellow and the other green ball. If the weights are equal, then the red and the yellow differ in weight. A weighing between these two balls allows us to deduce the weights of all other balls. If the red-green combination was heavier than the yellow-green combination in the first weighing, then the green ball in the red-green combination is certainly heavy and the other green is light. Now take the red from the red-green combination and the yellow from the yellow-green combination. Weigh these together against the remaining red and the remaining yellow. The only interesting case is when this weighing is “equal”. Then, the red from the red-green combination must be heavy and the yellow from the yellow-green combination must be light.
