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You have six balls. Two are red, two are white, and two are blue. One ball of each color weighs 6kg. The other three all weigh 6.01 kg.

How many uses of a two pan balance does it take to determine the weights of all the balls?

Source: Mathematical Circus, Martin Gardner, p. 127.

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You need

two weighings.

First,

weigh R1+W1 against R2+B2.

Now

if they balance, then we have either R:1 W:2 B:2 (meaning red ball 1 is the heavy one, etc.) or R:2 W:1 B:1. Weighing pretty much anything against anything else will distinguish these two.

On the other hand,

if one side is heavier, WLOG it's R1+W1 > R2+B2. Then we necessarily have R:1, and the others can't be W:2 B:2. So we have one of W:1 B:1, W:1 B:2, W:2 B:1. Now weigh W1 against B1. In those three cases we have respectively W1 = B1, W1 > B1, W1 < B1.

Now

no third weighing is necessary and this paragraph is just here so it isn't too obvious from the structure of my answer what the desired number is

and

I suppose I should say explicitly that if the first weighing yields R1+W1 < R2+B2 instead, we just need to swap "white" and "blue" and "light" and "heavy" in the second-weighing strategy above

and therefore

we are done, in all cases, after at most two weighings.

I believe

the first weighing here is essentially the only possibility that makes two weighings enough.

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Just a very slight variation to Gareth's correct answer and how I arrived at essentially the same answer

if you end up with R1+W1 > R2+B2, weigh R1+R2 against W1+B2. Because we know W1 is either greater or equal to B2: if R1+R2 > W1+B2 both W1 and B2 are light, if R1+R2 < W1+B2 then both are heavy and if R1+R2 = W1+B2 then W1 is heavy and B2 is light.

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