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 this is easy.. by me)

wops, i made a mistake, its 11 not 10 coins.. anyway its about the same solution... divide the coins into piles of 5, leaving 1 out. weigh the 2 piles against each other, if one is heavier, it contains the odd coin. if both weigh the same, the coin you left out is the odd one. lets say the odd coin is in a pile. divide this one into piles of 2 and leave 1 out once again. weight them; if one is heavier, weigh 1 coin against the other. if both weigh the same, the coin you left out is the odd one. as for the open problem, im not sure..

Posted by me on 2003-06-05 23:05:41