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.

i have driven myself insane with these sorts of problems. you can do 12 coins with three weighings. 39 with four. 120 with 5. someone try and guess whow many you can do with 6 weighings. (sidenote, i have the answers for all of these, up to and including 120 with five, talk to me at some other point if you wish to hear these

Posted by Jonathan on 2003-03-17 15:00:19