12 coins with a scale you get three tries. I think I remember how this works.
Six coins on each side the fake one is lighter thus the six real coins will weigh more so those six are real. Next weigh is three coins per side once again the three real ones weigh more so you have three left. Place two of the three on the scale If the scale is even both are real and last coin is the real one if one side weighs more the fake one is the lighter coin. Three tries of weight one coin found.
I think KN has it. The riddle revolves around weight since the only tool is the scale. Without the false one weighing differently there would be no way to pick it out since it's appearance is the same. That's my opinion anyways.
Edited: Kyrroeth on 26th Sep, 2018 - 3:26am
No, Paul is right. As Velon hinted at in his rambling internal monologue, not knowing if the false coin is heavier or lighter makes it much more complex. Instead of distilling down to a group of 3 for the final weigh as KN suggested, which would work if we knew heavier vs lighter, you have to distill it down to two coins by the second weigh in such that the third weigh in can be against a neutral coin you have already determine is just the normal weight. The complexity is in weigh in number 2 after starting with three piles of four. KN’s logic for his final weigh in CAN be applied to the first weigh in, in that you can learn something about the third pile despite it not being on the scale.
I don’t think Velon would be able to get over this hurdle, however, so I posted him being stuck right about where KN is at.
Maybe I'm just dense. I didn't think of that aspect of it. I have no idea how to come up with the answer. Not without knowing if it is heavier it lighter. The second weigh in is the problem. Following Kn's method would get you to a 50/50 guess basically. Good luck with this one *smile*.