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.

Unfortunately, this solution requires an extensive description...
Divide the 12 coins in three groups of four coins each: A, B and C. Then start weighing group A against B. Now the following two situations are possible:
(1) The A against B weighing is equal:
This means that one of the coins in group C has a different weight. Now take two of the coins in group C (C1 and C2) and weigh them against two coins of group A (A1 and A2, of which you know they have a correct weight). Again, there are two possible results:
(1a) C1 + C2 are as heavy as A1 + A2:
This means that C3 or C4 is the coin with a different weight, now you can determine which one it is by weighing for instance C3 against A1 (= a correct coin).
(1b) C1 + C2 differ in weight from A1 + A2:
This means that C1 or C2 is the coin with a different weight, so now you can determine which one it is by weighing for instance C1 against A1 (= a correct coin).
This doesn't seem to let me post really long message, so continued in next msg.

Posted by Half-Mad on 2002-05-16 02:19:28