You have nine brass rings, but three are actually gold. Can you pick these out using a balance scale three times at the most?

(In reply to re: The answer is ... by Tiralmo)

If you devise a strategy where what to weigh depends on the results of the previous weighing(s) it still is true that for any one weighing, the results can be either L, R or =, indicating the left pan is heavy, the right pan is heavy or the pans balance.

In the case of dependent weighings, the sequence of results, say LL=, would carry with it the type of the second weighing (by the initial L) and the type of the third weighing (by the previous LL).  Likewise with any other sequence of observed balances and non-balances.

So you're still limited to the possible sequences: LLL, LLR, LL=, LRL, LRR, LR=, L=L, L=R, L==, RLL, RLR, RL=, RRL, RRR, RR=, R=L, R=R, R==, =LL, =LR, =L=, =RL, =RR, =R=, ==L, ==R, ===, where, say RLR indicated the first weighing resulted in the right pan having the heavier set and the second weighing was done in accordance with what would be done then; and that second weighing resulted in the left pan having the heavier set; then the experimenter determines, based on these, what to weigh next, and in that weighing found the right pan's contents to be heavy.  There are only 27 of these.

Posted by Charlie on 2005-06-24 01:44:35