There is an old riddle: "You have twelve coins and a balance scale. One of the coins does not weigh the same as the other eleven, but you don't know if the odd coin is heavier or lighter than the others. How can you determine, in three weighings, which coin is the odd coin?"
With three weighings you can actually find the odd coin in a group of thirteen coins. Every version of this question that I have seen asked about a group of twelve coins, but the maximum is thirteen.
I wrote a computer program to find the minimum number of weighings for thirteen and larger numbers of coins; this was basically a turning point when I realized that I have no life.
Some versions of this problem say that at the end of the three weighings, you have to know which coin is the odd coin and know whether it is heavier or lighter than the others. In that case, twelve coins really is the maximum number of coins that can be solved in three weighings.
If the two pans are even, then you have five coins left to look at, and you still don't know whether the odd coin is heavier or lighter. Take three of those unknown coins, and three coins from the group of eight that you know are "normal", and weigh those groups of three against each other. If the unknown group of three is heavier, then you know that the coin is one of those three and you know that it's heavier than the rest, so you can easily find it in the last weighing. Similarly if the unknown group of three is lighter.
If the scales balance, then you know the odd coin is one of the last two coins. Take one of those coins and weigh it against one of the ten coins that you know are normal. Then if the scales don't balance, that coin is the odd coin, and if they do balance, the odd coin is the one left over.
Note that in the very last case -- "the odd coin is the one left over" -- you don't know whether the odd coin is heavier or lighter than the rest, since it never touched the balance scale. This is why if you want to find the odd coin in a group of thirteen and determine whether it's heavier or lighter than the rest, then you need four weighings; for that problem, the largest number of coins that can be solved in three weighings is twelve.