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 a better open problem by theBal)

I am not 100% sure (so please explain if I am wrong), but I don't think you can find the odd coin out of 9 coins in only 2 weighings. Not unless you know the odd coin is definitely heavier (or definitely lighter). But in that case, g(3) would be 27, wouldn't it?

Posted by nikki on 2003-03-06 09:32:27