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.

Well, we can easily figure out a solution with 12 coins if we know wether the odd one is lighter or heavier than the rest.

But I'm yet to think of a way to solve the problem without this info.

Posted by levik on 2002-05-03 07:21:36