In a pile, there are 11 coins: 10 coins of common weight and one coin of different weight (lighter or heavier). They all look similar.

Using only a balance beam for only three times, show how you can determine the 'odd' coin.

Open problem (i cannot solve this myself): how many more coins (with the same weight as the ten) can we add to that pile so that three weighing still suffices? My conjecture is zero, though my friend guessed that adding one is possible. The best bound we can agree upon is < 2.

(In reply to re: this is easy.. by me)

How do you conclude that the heavier pile contains the odd coin? The problem only states that one of the eleven coins has a different weight (lighter or heavier).

Posted by Sanjay on 2003-06-06 14:16:37