A flash, a thud, and Kinesys lands in a Staterion playground. Here even the swings have counterweights calibrated to the gram, and instead of hide-and-seek, the kids play guess-the-weight.
A little Staterian walks up with four identical marbles and a toy two-pan balance. “One weighs different from the others,” he explains, dead serious. “Whoever finds it in the fewest weighings wins.”
Dord is offended: nobody has ever challenged him to such an underrated game. Liz laughs and, while she’s at it, tries a pirouette on the swing. M00N, meanwhile, lines up eagerly: “Me! I want to play!”
The Riddle
You have four marbles of the same size, but one of them has a different weight from the others. You have to find this marble using a two-pan balance. What is the minimum number of weighings needed to solve the problem?
Hint
You don’t know whether the odd marble is heavier or lighter, and you don’t need to find out. A marble that balances against another one can serve as your reference.
Solution
Two weighings are enough. Let’s call the marbles A, B, C, and D.
In the first weighing, compare A and B, and note whether they balance. In the second weighing, remove B and replace it with C, that is, compare A and C, and again note whether they balance. Depending on the results of the two weighings there are four cases, and in each one the marble with the different weight is identified:
| Weighing 1 (A vs. B) | Weighing 2 (A vs. C) | Odd marble |
|---|---|---|
| Balanced | Balanced | D |
| Balanced | Unbalanced | C |
| Unbalanced | Balanced | B |
| Unbalanced | Unbalanced | A |
A single weighing wouldn’t be enough: with four candidate marbles, one comparison can’t tell them all apart.