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We have two white, two red and two blue 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?
Explanation :
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Two weighings suffice. Weigh one red and one white ball against one blue
and the other white ball. If the weights are equal, then the red and
the blue differ in weight. A weighing between these two balls allows us
to deduce the weights of all other balls. If the red-white combination
was heavier than the blue-white combination in the first weighing, then
the white ball in the red-white combination is certainly heavy and the
other white is light. Now take the red from the red-white combination
and the blue from the blue-white combination. Weigh these together
against the remaining red and the remaining blue. The only interesting
case is when this weighing is "equal". Then, the red from the red-white
combination must be heavy and the blue from the blue-white combination
must be light.
