Overview
Counterfeit-coin puzzles are a staple of recreational mathematics. You have a set of identical-looking coins, exactly one of which is off-weight, and a two-pan balance that only tells you which side is heavier — not by how much.
The key is that each weighing has three possible outcomes — left heavier, right heavier, or balanced — so a single weighing can split the suspects into three groups at once.
How to solve The Counterfeit Coin
- Divide the 8 coins into three groups: 3, 3 and 2.
- Weigh the two groups of three against each other.
- If one side is heavier, the fake is among those three; if they balance, it's in the group of two set aside.
- Second weighing: take two suspects and weigh them against each other. The heavier one is fake — or, if they balance, the untested suspect is fake.
The key insight
Because each weighing has three outcomes, k weighings can distinguish among 3^k cases. Two weighings cover up to 9 coins — so 8 is comfortably solvable, and this is why you divide by three, not two.
Variations & echoes
- The classic hard version has 12 coins where you don't even know if the fake is heavier or lighter — still solvable in three weighings.
- The 'divide into three' idea is exactly ternary search, the base-3 cousin of binary search.
- Information theory sets the limit: you need at least ⌈log₃(cases)⌉ weighings.
Frequently asked questions
Why divide into three groups, not two?
Because a balance gives three outcomes, dividing into three uses every weighing to its fullest — halving would waste the 'balanced' result.
Could you do 9 coins in two weighings?
Yes. Since 3² = 9, exactly nine coins is the maximum solvable in two weighings when you know the fake is heavier.