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 you can do up to 27 with 3 times weighing by Ali)

No I don't think It's that easy. Your method would work if you knew that the fake coin's weight was lighter or heaver than the other ten. But you don't know. So on your first step you messed up.

When you weigh the two piles of the same amount of coins and their weights aren't the same, the fake coin could be in te lighter side (It being lighter) or on the heavier side (It being heavyer). So you'd have to know if it was heavier or lighter first.

Posted by ron on 2004-12-31 16:53:07