This is discussion archived from a time before the current discussion method was installed.
Iron Salticus: In the page, it says "Thirteen coins and the unknown weight is not possible in the standard three weighings", but I believe this to be false; unless I am mistaken the following method should work:
First, put four coins on each side. If it is unbalanced, then solve it as you would in the twelve-coin version. If it is balanced, then you have cleared eight coins, and have five left. Call them A, B, C, D, and E, and call each the other coins Q.
Next, put A and B on one side of the balance, and put C and Q on the other side. This is the second weighing. If it is balanced, then simply weigh D against Q; if it balances, E is the odd weight, and if it is unbalanced, then D is the odd weight. If the AB/CQ weighing is unbalanced, then your third weighing is instead A against B. If it is unbalanced, then which ever coin has the same state as AB had in the second weighing is the odd weight (so if AB was too heavy, the heavier coin of A and B is odd, and if AB was too light, then the lighter coin is odd). If A/B is balanced, then obviously C is the odd weight. Regardless, you can determine which one is unusual.
Can someone confirm whether or not this logic is valid?
Vilui: It's not valid. In the case where AB/CQ balances and D/Q balances, you know E is the odd weight but not whether it's heavier or lighter.
In fact (and I apologise for a comment stating the opposite I made earlier) 13 coins definitely is impossible. Proof: the first weighing must have an equal number of coins on each side, or it tells you nothing. If it's four against four (or fewer) and it balances, there are ten (or more) possibilities for which of the remaining coins is the odd one out and whether it's heavier or lighter. But the remaining two weighings can only distinguish nine possibilities with certainty, since there are three possible outcomes for each weighing. If instead the first weighing is five against five (or more) and it doesn't balance, again you have ten (or more) possibilities remaining.
