Question: You are give two red balls, two green balls and two yellow balls. One ball from each color is heavier than the other one. All the heavier balls weigh the same and all the lighter balls weigh the same. You are only provided with a weighing balance. How many tries would it take to identify which one is heavier and which one is lighter?

Answer: Let’s label the balls R1, R2 (Red ones), G1, G2 (Green ones) and Y1, Y2 (Yellow ones).

First weigh R1, G1 on one side and R2, Y1 on the other.

If they are equal, we know that one of G1 or Y1 is heavy. We can just weigh them both and find out which one is heavier. If G1 is heavy, the heavier set is {G1, R2, Y2} and the others are lighter. If Y1 is heavy, the heavier set is {G2, R1, Y1}.

If R1, G1 is heavy, we know that either G1 is heavy or Y1 is light. We proceed to weigh G1, Y1 with G2, Y2. If they are equal, G1 is the heavy one. The heavier set is {G1, Y2, R1}. If G1, Y1 is heavy, G1 and Y1 are both heavy. The heavier set is {Y1, G1, R1}. If G2, Y2 is heavy, G2 and Y2 are both heavy. The heavier set is {R1, G2, Y2}.

If R2, Y1 is heavy, we know that either Y1 is heavy or G1 is light. This is similar to the case above. Try to work it out yourself before continuing with the solution. Weigh G1, Y1 with G2, Y2. If they are equal, Y1 is heavy. The heavier set is {Y1, R2, G2}. If G1, Y1 is heavy, G1 and Y1 are both heavy. The heavier set is {G1, Y1, R2}. If G2, Y2 is heavier, G2 and Y2 are both heavy. The heavier set is {G2, Y2, R2}.

Therefore, in any of these cases, we only need two tries with the balance.

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