Logic Puzzles at quizzes-online.com

To make the problem extra challenging, you can only make 3 weighings plus you're asked to determine if the counterfeit coin is heavier or lighter than the legitimate coins.

Further thoughts: The solution I've given takes different paths depending on the outcomes of previous weighings. However, by tweaking my answer you should be able to devise a solution where each weighing is independent of the previous weighings.

Hints will appear here!

There are many possible solutions and approaches to this problem. The one I'll describe isn't the simplest, but is perhaps one of the more "obvious" solutions in that it arises naturally by thinking of the possible moves one could make.

Start by thinking of your first move. The only sensible thing to do is to measure an equal number of coins on each pan of the balance. Think about the two possible kinds of outcome.

Think about the end game. Suppose you've already determined whether the counterfeit coin is lighter or heavier than the other coins. Now imagine that you're left with exactly three suspect coins. You should be able to determine in one weighing which coin is the counterfeit.

Given this information, you should make it your goal to ensure that after the second weighing you know:

That the counterfeit coin is in a particular set of (at most) three coins, and
Whether the counterfeit coin is lighter of heavier than the other coins.

I'll number the coins from 1 to 12. In the first weighing we perform:

1 2 3 4 with 5 6 7 8

If they balance, then the fake coin is amongst the remaining four coins. In this case, perform:

1 2 3 4 with 9 10 11 5

Now if the fake is in the set {9 10 11} we can find it in one weighing, as in Hint 2, else we know 12 is the fake and it's easy to determine whether it's heavy or light in one weighing.

Suppose instead that the first weighing didn't balance. Without loss of generality, we'll assume that {1 2 3 4} is heavier than {5 6 7 8}. This time we weigh:

1 2 3 8 with 9 10 11 4

If the left pan is lighter then 8 is the fake and is light. If they balance, then the fake is in {5 6 7} and is light. If the left pan is heavier then the fake is in {1 2 3} and is heavy. In either of these last two instances, we can determine the counterfeit coin with one more weighing.