Counterfeit Coin Puzzle Solver
Find the false coin in three weighings, and see why the puzzle says twelve when the counting bound says thirteen — a parity argument, not caution.
Three weighings handle thirteen coins, not twenty-seven
A balance has three outcomes, so three weighings give twenty-seven results and the obvious guess is twenty-seven coins. The real answer is thirteen, and the gap is worth understanding because it is not arbitrary — it is (3n − 1) ⁄ 2, and each part of that expression is doing a specific job.
| Weighings | Results | Coins you can handle |
|---|---|---|
| 1 | 3 | 1 |
| 2 | 9 | 4 |
| 3 | 27 | 13 |
| 4 | 81 | 40 |
| 5 | 243 | 121 |
The minus one is easy: every coin balancing cannot be an answer, because one coin definitely is false. The division by two is the interesting half. Think of each coin as a plan — for each weighing it goes left, right, or stays out. The sequence of tips you observe simply is that plan, negated if the coin turns out to be light rather than heavy. So a heavy coin and a light coin whose plans are exact mirrors read identically at every weighing, forever. Twenty-six usable plans pair off into thirteen coins.
So why does the puzzle always say twelve?
Not out of caution. Thirteen genuinely cannot be done with only the suspect coins, and the reason is parity rather than a failed search. To handle the full (3n − 1) ⁄ 2 coins you must use every plan exactly once — there are no spares. So how many coins sit on the pans is no longer a choice you get to make:
| Weighings | Coins at capacity | On the pans, each weighing | |
|---|---|---|---|
| 1 | 1 | 1 | odd — cannot split evenly |
| 2 | 4 | 3 | odd — cannot split evenly |
| 3 | 13 | 9 | odd — cannot split evenly |
| 4 | 40 | 27 | odd — cannot split evenly |
| 5 | 121 | 81 | odd — cannot split evenly |
| 6 | 364 | 243 | odd — cannot split evenly |
That column is always 3n−1, and 3n−1 is always odd. An odd number of coins cannot be divided evenly between two pans, and an uneven weighing tells you nothing at all. Which side you put each coin on cannot change it, because flipping a plan leaves its blanks exactly where they were. So full capacity always needs one coin you already trust — at every size, not just at thirteen.
| Coins | Using only the suspect coins | With one coin you already trust |
|---|---|---|
| 12 | possible | possible |
| 13 | impossible | possible |
| 14 | impossible | impossible |
There is a floor as well, and it fails for the opposite reason. The workable range at three weighings is exactly 3 to 12 coins. One coin cannot be weighed against anything. Two coins can only go one per pan, where their plans are exact mirrors and no number of weighings will ever separate them. Too many coins and the pans will not balance; too few and the answers collide.
It is easy to get this wrong in the other direction too. A search that only ever weighs coins still under suspicion — treating cleared coins as spectators rather than counterweights — cannot reach thirteen at all, and will report twelve as if it were the limit.
You can write all three weighings down in advance
The strategy above is non-adaptive: the second weighing does not depend on how the first came out, so you could hand the whole plan to someone else and just read the three answers off at the end. That is a stronger result than the puzzle asks for — the usual solution branches after each weighing — and it falls straight out of thinking about coins as plans rather than as a decision tree.
How to use
- Choose how many coins are on the table.
- The three weighings are listed in advance — no branching required.
- Record what the balance did at each one.
- The false coin is named, along with whether it is heavy or light.
Frequently asked questions
How many coins can three weighings handle?
Thirteen, if you have one coin you already know is genuine. Twelve if you do not. The general limit is (3 to the power n, minus one, all over two) — so 1, 4, 13, 40 and 121 coins for one to five weighings.
Why is it not twenty-seven coins for three weighings?
Two reasons. The all-balanced result cannot mean anything, since one coin definitely is false, which removes one. And a heavy coin reads exactly like a light coin following the mirrored plan, which halves the rest. Twenty-seven becomes twenty-six becomes thirteen.
Why does the puzzle always use twelve coins rather than thirteen?
Because of parity. Handling the full thirteen forces you to use every possible plan, which puts nine coins on the pans at each weighing — an odd number, which cannot be split evenly between two sides. An uneven weighing tells you nothing, so you need one extra coin you already trust to even up a pan.
Is that true at other sizes as well?
Yes, at every size. At capacity the number of coins on the pans is always 3 to the power n minus one — 1, 3, 9, 27, 81, 243 — and every one of those is odd. So the largest pile you can sort out using nothing but the suspect coins is always one short of the counting bound.
What is the smallest number of coins this works for?
Three, without a reference coin. One coin cannot be weighed against anything at all. Two coins can only go one per pan, where they are exact mirrors of each other and no number of weighings will ever separate a heavy first coin from a light second one.
Do you have to decide the second weighing after seeing the first?
No, and that is a stronger result than the puzzle asks for. The strategy here is non-adaptive: all three weighings are fixed in advance, so you could hand the whole plan to someone else and just read the three answers off at the end.
Does it tell you whether the coin is heavy or light?
Yes, both together. That is what makes the puzzle hard — if you already knew the fake was heavy, three weighings would handle twenty-seven coins rather than thirteen, because the mirror collision that halves the count would not arise.
Does this send anything anywhere?
No. The strategy is searched for in your browser and nothing is uploaded.
🔒 This tool runs entirely in your browser. Nothing you enter is uploaded, logged, or stored.