Golden rings  logic puzzle
A jeweller has three boxes: A, B and C. Each box contains three rings, which are either all made of pure gold, or all fake. The fake rings weight 1 gram less then the real ones. It is known that a least one of the boxes contains fake rings. The jeweller has a balance with three weights of exactly 1 gram. Is it possible that the jeweller finds out which box(es) contain the fake rings with weighing only once? And if so, how does she do it? (We define weighing only once that only once a situation is reached in which the balance is in equilibrium.)
Explanation
On the lefthand side of the balance, the jeweller puts 1 ring from box A and 2 from box B. On the righthand side she puts three rings from box C. If equilibrium is reached, all rings are fake.
If equilibrium has not yet been reached, then there are six possibilities to reach equilibrium. If 1 gram is needed on the lefthand side, then only the rings from box A are fake.
If 2 grams are needed on the lefthand side, then only the rings from box B are fake.
If 3 grams are needed on the lefthand side, then only the rings from box A and B are fake.
If 1 gram is needed on the righthand side, then only the rings from box B and C are fake.
If 2 grams are needed on the righthand side, then only the rings from box A and C are fake.
If 3 grams are needed on the righthand side, then only the rings from box C are fake.
If equilibrium has not yet been reached, then there are six possibilities to reach equilibrium. If 1 gram is needed on the lefthand side, then only the rings from box A are fake.
If 2 grams are needed on the lefthand side, then only the rings from box B are fake.
If 3 grams are needed on the lefthand side, then only the rings from box A and B are fake.
If 1 gram is needed on the righthand side, then only the rings from box B and C are fake.
If 2 grams are needed on the righthand side, then only the rings from box A and C are fake.
If 3 grams are needed on the righthand side, then only the rings from box C are fake.
