You have a unique three-pan balance with which to measure the weight of coins. This balance has three pans, as its name implies. Each pan can be loaded with any number of coins after which you may press a button to turn the balance on. If all three pans have different weights, the scale will indicate to you the pan that has the 2\(^\text{nd}\) heaviest weight (so the pan that is not the heaviest nor lightest). However, if any two pans have the same weight, then the scale will read "ERROR" (because there is no unique 2\(^\text{nd}\) heaviest weight). After that, the balance switches off, and you can unload and reload the pans again for the next use.

In a pile of 64 coins, all but one of them have the same weight; the odd one is heavier. You want to identify this heavier coin in as **few button presses** (uses of the balance) as possible. With the best strategy, how many button presses do you need in the worst case?

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