You have six balls - three red and three black. All the red balls weigh differently, i.e. one of them is heavy, the other medium, and the third light. Each red ball has a black twin of the same weight. A heavy and a light ball put together weigh as much as two medium balls.
What is the least number of weighings required on a balancing scale to determine which is which?
Make the first weighing one red vs one red. Call the heavier of the two R1 and the lighter R2. Call the red ball left off the scale R3.
Make the second weighing one black vs one black. Call the heavier of the two B1 and the lighter B2. Call the black ball left off the scale B3.
Make the third weighing R3 vs B3. From here there are three cases:
Case 1: R3>B3
R2 must be light and B1 must be heavy. The remaining four balls in the order {R1, R3, B2, B3} must be one of three possibilities: {heavy, medium, medium, light}, {medium, heavy, medium, light}, or {medium, heavy, light, medium}. Making a fourth weighing of (R1+B2) vs (R3+B3) or (R1+B2) vs (R2+B1) will determine which of the three possibilities is right.
Case 2: R3<B3
B2 must be light and R1 must be heavy. The remaining four balls in the order {B1, B3, R2, R3} must be one of three possibilities: {heavy, medium, medium, light}, {medium, heavy, medium, light}, or {medium, heavy, light, medium}. Making a fourth weighing of (R2+B1) vs (R3+B3) or (R1+B2) vs (R2+B1) will determine which of the three possibilities is right.
Case 3: R3=B3
All these equalities are true: R1=B1, R2=B2, and R3=B3. Make a fourth weighing of (R1+B2) vs (R3+B3) or (R2+B1) vs (R3+B3). With R1>R2 already known, this will determine the weights of each equal pair.
Another Solution
