[emilfihlman]

How do you solve this?

ABC, one of them is different weight than the other two (either more or less)

Possible ways to weigh them:

  A v B
  A not enough information
  E C is odd one out
  B not enough information

  A v C
  A not enough information
  E B is odd one out
  C not enough information

  B v C
  B not enough information
  E A is odd one out
  C not enough information

  AB v C
  AB not enough information
  E C is odd one out
  C C is odd one out

  AC v B
  AC not enough information
  E B is odd one out
  B B is odd one out

  BC v A
  BC not enough information
  E A is odd one out
  A A is odd one out

In all possible ways to weight 3 coins, all have uncertainty which means you cannot deduce which is the odd one out.

[scrooched_moose]

Your OP stated one is lighter, not one is different.

It's easier than the 12-coin one different puzzle, but introduces the ternary concept well enough to get you started on the harder problem (there are some other complications to it as well).

Weigh 2 coins. If the scales move you've found the lighter coin. If the scales balance, it's the third coin you didn't weigh.

[emilfihlman]

Ah I got the original (12 coins, one is lighter or heavier) mixed up with 3 coins (one is lighter).

[grive]

Label each coin, divide in three stacks each time, and shuffle meaningfully each time the subsets, to remove uncertainty.