If we assume the classic balance puzzle (Twelve balls and a scale) is phrased like this:

You have 12 coins that are identical except for one counterfeit that is heavier or lighter. Can you use a balance scale three times to determine the counterfeit coin and say if it was heavier or lighter?

If "balance scale" was specifically a 2-pan balance whose arms had adjustable length, can you solve the above puzzle with 13 starting coins?

The idea is that before each weighing, you can adjust either arm length however you want, but not during. The weights of the arms themselves (or the pans) do not influence your outcome (in reality you would have to account for that stuff). I think many people will understand what all this is getting at, but in case you don't appreciate the physics or terminology read the below spoiler:

For a "standard" balance scale, each pan has an equal-length arm. The arm sets the distance between the pan and the pivot point of the scale. This means that both pans "fight" each other with equal strength. If I make the left arm 3 times longer than the right arm, then the left side fights the right side with 3 times the strength. Why is any of that important? You can, assuming you set your arm lengths appropriately, compare any positive number of coins on one side to any positive number of coins on the other, and the balance will still behave like you are used to. A heavy counterfeit coin still causes its pan to go down and a light counterfeit coin still causes its pan to go up.


1 Answer 1


I take no credit for the answer, as it is entirely Joe Z. 's brainchild. https://puzzling.stackexchange.com/a/224/48794

As he mentions himself, there's no LLL and RRR because you can't have 9 balls on the balance. However, with the movable arms, you can! By having the ratio of the arm lengths be 4:5, you can fit an extra ball on the left, allowing you to squeeze in the LLL ball!

  • $\begingroup$ Well, even if you were standing on the shoulders of giants, I can at least give you credit for putting the pieces together so fast. $\endgroup$
    – Skosh
    Jan 26, 2019 at 16:28

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