Solve the Puzzle: Find the Odd Coin in 3 Weighings

[Harut82]

You have twelve coins, eleven identical and one different. You do not know whether the "odd" coin is lighter or heavier than the others. Someone gives you a balance and three chances to use it. The question is: How can you make just three weighings on the balance and find out not only which coin is the "odd" coin, but also whether it's heavier or lighter?

 

[GCT]

this has been posted here repeatedly. There's a variation to this coin problem that is much more complex, I saw it at "ibm ponder this" a couple of years ago...it makes this problem seem too easy. You can search there "ponder this" database if you want a real challenge.

 

[Architeuthis Dux]

This is a classic problem in the field of puzzle-solving and mathematical logic. In order to solve this puzzle in three weighings, we need to use a strategy called "divide and conquer". Here is one possible solution:

1. First, divide the twelve coins into three groups of four coins each. Place one group on each side of the balance.

2. If the two sides of the balance are equal, then the odd coin is in the fourth group that was not weighed. If one side is heavier than the other, then the odd coin is in that group.

3. Now, take the group with the odd coin and divide it into two groups of two coins each. Place one group on each side of the balance.

4. If the two sides of the balance are equal, then the odd coin is one of the two coins that were not weighed. If one side is heavier than the other, then the odd coin is in that group.

5. Finally, take the group with the odd coin and weigh the two coins against each other. If one side is heavier, then that is the odd coin. If they balance, then the odd coin is the one that was not weighed.

This strategy works because each weighing provides us with information that helps narrow down the possible locations of the odd coin. By dividing the coins into smaller groups, we are able to eliminate more possibilities with each weighing. This approach can be applied to any number of coins as long as we have three weighings to use.