How do you know how many coins to start off with on each side of a "find the counterfeit coin using a 2-pan weigh scale" problem? The fake coin is defined by either having a lesser or greater weight than all of the other coins in the problem.
Say there are 12 coins, 1 out of the 11 which is the fake coin. How does everyone know to start off by initially weighing 4 coins against 4 coins? Why not 6 against 6 (since there are 12)? Why not 1 against 1, 2 against 2, or 3 against 3 either? Why is it general consensus that 4 coins against 4 coins is the optimal initial weighing?
Same thing if the problem were to have 8 coins, 1 of which were the fake. How does everyone conclude to initially weigh 3 coins against 3? Why not 4 against 4, 1 against 1, or 2 against 2?
There must be logic behind determining the initial weighing?
 A: For each weighing, there are three possible outcomes.  Thus, N weighings can yield $ 3^N$ outcomes.
The fake coins is one of 12, and it can be heavy or light (2 possibilities), so you have a total of 24 possible answers.  You must break that into 3 groups of not more than 9 each in the first weighing.
If you put 6 on each side, then you have zero possible answers for the scale to come up balanced, which means 24/2 = 12 for each direction, left or right.  12 is more than you can differentiate in two more weighings, so that won't work.  Similarly, if you put 5 on each side, there are 4 possible answers where the scale is even and (24-4)/2 = 10 for the scale to tilt either direction.  10 is also too many.
If you put 4 on each side, that leaves for unweighted, so there are 4x2=8 possible answers where the scale is balanced and (24-8)/2=8 possible answers for it to tilt either direction.  These are all $\leq$ 9, so each of them should work.
Then, for each combination, you need to break up the 8 possible answers to three groups of not more than 3 each, so your third weighing can find the final answer.
This general logic: break the set of answers up into 3 groups of nearly equal size, is the path to solutions for almost all weighing problems.
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