Originally posted by murrow apologies if this has appeared before

Indeed it has. In fact, I was thinking of compiling a 'Posers & Puzzles FAQ' section on my website, including a general approach to the balance problem. In this case, the answer is that 2 weighings are needed, but I'll leave others to explain why.

okay-3 against 3. if they balance, weigh the remaining two. else, weigh 2 from the lightest side against each other. if they balance, it's the other one from that side. and if they don't, the lightest one is -well- the lightest one! ðŸ˜€

3*3
if equle, then the other two for an answer.
if one is less, then of the lighter, 1*1
if equle, then the one left out
if one is lighter, then that one.