The Twelve Coins
This is the hardest and most famous weighing puzzle there is. If you have read The Lighter Coin (puzzle #1) and its solution, you have exactly the tool you need — and you will still find this one difficult.
There are twelve coins. Eleven are identical. One is different: it might be heavier, or it might be lighter, and you are not told which.
You have a two-pan balance again — three outcomes, no numbers.
Twelve coins. One of them is wrong in an unknown direction.
The difficulty in one picture: a tilt does not tell you which side is lying.
It really can be done, every time. Work out how, then count what the method is worth: each weighing still has three outcomes, but now every coin has two possible faults, and one outcome has to be spent on the case where nothing is wrong.
Answer to type: with four weighings instead of three, what is the largest number of coins you could handle under these same rules — finding the odd coin and saying whether it is heavy or light?
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Count the possible answers, not the coins. With n coins there are 2n possible situations, and n weighings can tell apart 3ⁿ things — but not every one of those outcomes is usable.
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Answer: 39
First, that three weighings really do settle twelve coins. Label them 1-12 and weigh 1,2,3,4 against 5,6,7,8.
- If they balance, the odd coin is among 9-12 and you have eight coins you know are genuine. Weigh 9,10,11 against three genuine coins: if it balances, coin 12 is the odd one (weigh it against a genuine coin to learn heavy or light); if it tips, you know both the group of three and the direction, and one more weighing of 9 against 10 finds it.
- If they tip — say the left side sinks — then either one of 1-4 is heavy or one of 5-8 is light. Eight suspects, each with a known direction. Now weigh 1,2,5 against 3,4,6, mixing the two groups deliberately. Whichever way this goes, only two or three suspects survive, and the third weighing separates them.
- So twelve coins, three weighings, direction included. Now count why twelve is the limit.
- With n coins there are 2n possible answers: any coin could be the odd one, heavy or light. Three weighings give 3 × 3 × 3 = 27 distinguishable outcomes.
- But three of those outcomes are unusable: if every weighing balances you learn nothing, and the balance can never distinguish "all coins genuine" cases. The usable count is 27 − 3 = 24, and 2n ≤ 24 gives n ≤ 12. Exactly twelve — the puzzle is tight to the bone.
- For four weighings: (3⁴ − 3) / 2 = (81 − 3) / 2 = 39 coins.
The trick: counting answers instead of trying arrangements. Before you invent a method, ask how much information you need (2n possibilities) and how much your tool can supply (3ⁿ outcomes). That comparison tells you whether to keep looking for a method or to stop and prove it is impossible — and it is the founding idea of information theory.