Level
2

6
5
3
4

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This problem was created by my father Howard Grossman and published Scripta Mathematica, Vol. XI, in 1945. It subsequently appeared in a Scientific American column authored by his friend Martin Gardner. The answer is 3. There are two ways of achieving this result: the direct way, and my Dad's way.

Number the coins from 1 to 12. The direct way is to start by weighing 1234 against 5678. If they balance, then weigh 19 against 10,11 and the final step is obvious. If 1234 does not balance 5678 then weigh 125 against 346. If the imbalance is in the same direction as the first weighing, weigh 1 against 2; in the opposite direction weigh 3 against 4. But if 125 balances 346, then weigh 7 against 8.

My Dad's solution is to use the ternary number system: The first weighing is represented as -1,0,+1, i.e., count -1 if the left pan goes up, +1 if the right pan goes up, and 0 if the pans balance; the second weighing is represented as -3, 0, +3; and the final weighing is represented as -9, 0, +9. So1=1+0+0; 2=-1+3+0; ...; 9=0+0+9; 10=1+0+9; 11=-1+3+9; and 12=0+3+9. A + sign means the coin goes on the left pan; a - sign means it goes on the right pan; and a 0 means that coin is not part of that weighing. This tells you how each coin should be used in each weighing, i.e., surprisingly the choice of coins for the later weighings doesn't depend at all on the earlier results. There is a slight hitch because in the second weighing you need to switch one coin from one pan to the other to have the same number of coins on both pans, and in the third weighing you need to switch 4 coins. So arbitrarily, change the numbering of coins 9, 10, 11, 12 to -9, -10, -11, and -12 and then if one of them is the answer, you need to reverse lighter and heavier.