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 12 coins conclusive answer by John Smith)

13 coins can also be done.

Label the coins A1,A2,A3,A4,B1,B2,B3,B4,C1,C2,C3,C4,C5.

1st weighing:  A1,A2,A3,A4 vs. B1,B2,B3,B4

If this does not balance then the odd coin is one of those 8.  See John Smith's answer for details on determining the coin in that case.

If the odd coin is one of the C coins, then

2nd weighing:  A1,A2,A3 vs. C1,C2,C3

If the balance is even, then C4 or C5 are the odd coin and the third weighing will be between C4 and A1.  If balanced then C4 is odd.  If unbalanced then C5 is odd.

If the balance is uneven and since we know A1,A2,A3 are normal coins C1,C2,C3 are either heavier or lighter.  Since we now know that the odd coin is heavey (or light), we can have the third weighing between C1 and C2.  If they balance then C3 is odd.  If they are unbalanced then we use the knowledge of when the odd coin is heavy or light to determine whether C1 or C2 is the odd coin.

I believe 13 is the max as I couldn't figure a solution for 14 but I could be wrong of course.

Posted by Morgan on 2005-12-14 14:29:42