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Six Colored Balls

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.

Previous Puzzle: Treasure Island

An old parchment has directions to a treasure chest buried in an island:

“There is an unmarked grave and two tall oak trees. Walk from the grave to the left tree, counting the number of steps. Upon reaching the left tree, turn left by 90 degrees and walk the same number of steps. Mark the point with a flag. Return to the grave. Now, walk towards the right tree, counting the number of steps. Upon reaching the right tree, turn right by 90 degrees and walk the same number of steps. Mark this point with another flag. The treasure lies at the midpoint of the two flags.”
A party of sailors reached the island. They find a pair of tall oak trees merrily swaying in the wind. However, the unmarked grave is nowhere to be found. They are planning to dig up the entire island. It'll take a month. Can they do any better?
Next Puzzle: Josephus Problem

There are n persons in a circle, numbered 1 thru n. Going around the circle, every second person is removed from the circle, starting with person number 2, 4, and so on. Show that the number of the last person remaining in the circle can be obtained by writing n in binary, then moving the leftmost 1 to the right. So for example, with n = 13 persons (1101 in binary), the last person is number 11 (1011 in binary).

12 Sep 2008
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