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.

For the first weight: take 10 coins with one left over. Weigh 5 on each side. If sides are equal your left over coin is the odd one if not take the heavier side and keep.

For the second weight: take the 5 coins you kept and put 2 on each side with one left over. Again, if the sides are equal the left over is the one and if not keep the heavier side.

For the third weight: take the 2 coins you kept and weigh and put one on each side, the heavy side is your odd coin.

Posted by Jay on 2003-03-16 11:35:36