riddle of the day

A two-pan scale does not give numbers: it only says left, right or balance. The challenge is to squeeze these three answers to locate one coin out of twelve and know if it weighs more or less.

The fake coin among twelve

Strategist
Master plays

Riddle statement

You have 12 seemingly identical coins. One of them is fake, and you don't know if it weighs more or less than the others. You have a two-pan scale.

What is the minimum number of weighings necessary to always identify the counterfeit coin and know if it is heavier or lighter?

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Solution

Answer: the minimum is 3 weighings.

First let's see why 2 weighings cannot be enough. Each weighing has three possible results, so two weighings only allow us to distinguish:

$ 3^2=9 $

patterns of results.

But there are 24 cases to distinguish: any of the 12 coins can be false, and they can also be heavier or lighter. Therefore, 2 weighings are not enough.

With 3 weighings there are:

$ 3^3=27 $

possible patterns, so in principle there is enough information. Below is a strategy that achieves this.

Number the coins from 1 to 12.

First weighing: weighs 1, 2, 3, 4 against 5, 6, 7, 8.

Case 1: the first weighing is balance.

Coins 1 to 8 are authentic. The false one is between 9, 10, 11 and 12.

Second weighing: weigh 9, 10, 11 against 1, 2, 3.

  • If they balance, the false one is 12. Third weighing: weigh 12 against 1 to know if it is heavier or lighter.

  • If 9, 10, 11 weigh more, the fake one is between 9, 10 and 11, and is heavier. Third weighing: weighs 9 against 10. If one weighs more, that is the false one; If they balance, the false one is 11.

  • If 9, 10, 11 weigh less, the false one is between 9, 10 and 11, and is lighter. Third weighing: weighs 9 against 10. If one weighs less, that is the false one; if they balance, the false one is 11.

Case 2: the first weighing does not balance.

Suppose that 1, 2, 3, 4 weigh more than 5, 6, 7, 8. Then the false one is one of these eight possibilities:

  • one of 1, 2, 3, 4 —and it is heavier—, or

  • one of 5, 6, 7, 8 —and it is lighter.

The coins 9, 10, 11 and 12 are authentic.

Second weighing: it weighs 1, 2, 5 against 3, 6, 9.

  • If they balance, there are only three possibilities left: 4 heavy, 7 light or 8 light. Third weighing: weighs 7 against 8. If one weighs less, that is the false one; if they balance, the false one is 4 and is heavier.

  • If the left side weighs more, the compatible possibilities are 1 heavy, 2 heavy or 6 light. Third weighing: weigh 1 against 2. If one weighs more, that is the false one and is heavier; If they balance, the false one is 6 and is lighter.

  • If the right side weighs more, the compatible possibilities are 3 heavy or 5 light. Third weighing: it weighs 3 against a genuine coin, for example 9. If 3 weighs more, the fake is 3 and is heavier; If they balance, the false one is 5 and it is lighter.

If the opposite had happened in the first weighing - 5, 6, 7, 8 heavier than 1, 2, 3, 4 -, the same reasoning is applied, exchanging the sides and changing "heavy" to "light" where appropriate.

Thus, 3 weighings are always enough, and 2 cannot be enough.