12 coins, and one counterfeit.

[kant]

You have 12 coins. one of the coin is a counterfeit of some unkown weight. The rest of the coins have identical weight. Use three tries in a pan balance to identify the counterfeit.

 

[Lisa!]

Someone lese's posted it before!

 

[Architeuthis Dux]

First, I would divide the 12 coins into three groups of four coins each. I would then place two groups on each side of the pan balance, leaving the third group aside. If the two groups balance out, then the counterfeit is not in those two groups. If one side is heavier, then the counterfeit is in that group of four coins.

Next, I would take the group of four coins that contains the counterfeit and divide it into two groups of two coins each. I would again place one group on each side of the pan balance. If the two groups balance out, then the counterfeit is one of the two remaining coins. If one side is heavier, then the counterfeit is in that group of two coins.

Finally, I would take the group of two coins that contains the counterfeit and place one coin on each side of the pan balance. If the two coins balance out, then the remaining coin is the counterfeit. If one side is heavier, then that coin is the counterfeit.

By dividing the coins into smaller groups and using the pan balance to compare weights, we can narrow down the possible counterfeit coin to just one in three tries. This method is efficient and reliable in identifying the counterfeit coin among the 12 coins.