In a pile, there are 11 coins: 10 coins of common weight and one coin of different weight (lighter or heavier). They all look similar.

Using only a balance beam for only three times, show how you can determine the 'odd' coin.

Open problem (i cannot solve this myself): how many more coins (with the same weight as the ten) can we add to that pile so that three weighing still suffices? My conjecture is zero, though my friend guessed that adding one is possible. The best bound we can agree upon is < 2.

The problem is easily solved with threee weighings, considering two different cases where the odd coin does not lie in any trio and where the odd coin lies in a trio,

Edited on October 26, 2023, 8:54 pm

Posted by K Sengupta on 2023-10-26 20:53:13