Find the heavy ball among 27 in three weighings

Question

You have 27 balls that look identical, but exactly one is slightly heavier than the rest (which all weigh the same). You have a two-pan balance scale that tells you which side is heavier, or that the two sides match. What is the fewest weighings that guarantees you find the heavy ball, and how?

Accepted Answer

Three weighings always suffice, and you can't guarantee it in fewer. Why three is the floor. Each weighing has three possible outcomes: left heavier, right heavier, or balanced. So k weighings can produce at most 3^k distinguishable result-patterns. To pin down 1 ball out of 27 you need 3^k \ge 27. Since 3^2 = 9 < 27 and 3^3 = 27, you need 3 weighings (and 3 is exactly enough). The method: split into thirds each time. 1. Split the 27 into three groups of 9. Put one group of 9 on each pan; leave 9 aside. If a pan drops, the heavy ball is in those 9; if they balance, it's in the 9 set aside. Ei…

Find the heavy ball among 27 in three weighings

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