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Separate the Marbles...

There are three pairs of similar marbles of three different colors, Red Green and Blue.

In each pair, one is light and the other heavy. All light marbles are identical in weight and all heavy marbles are identical in weight.

The difference in light and heavy is miniscule. The problem is you have a weighing balance but no weights and measures.

If you were allowed three attempts at weighing, you could weigh three pairs three times separately and segregate light from heavy easily. But that's so mundane.

So your task is that you are allowed only two attempts at weighing and must still succeed in segregating the three pairs.

I found three distinct ways to do it. How many ways can you find?

Note by Satyen Nabar
2 years, 8 months ago

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Well, here's one way. Put, say, a red and a green marble together in one tray, and a green and a blue marble together in the other. The following scenarios can then occur:

(i) If the two trays balance, then take the red marble just used and weigh it against the other red marble. If the first red marble is heavier, then the green marble it was with in the last step was the light one and the blue marble from the first weighing was the heavy one. If the first red marble is lighter, then the green marble it was with in the first weighing was the heavy one, and the blue marble the light one.

(ii) If the tray with the red and green marble is heavier, then this green marble is the heavy one, and if the blue marble on the other side is the heavy one, then so is the red marble that was alongside the green one. If the blue marble on the other side is the light one, then the red marble that was alongside the green one could be either the light or heavy one. To resolve this, we then weigh the blue marble just used against the unweighed red marble. If they weigh the same, then they must both be the light ones. If the blue one is lighter, then the first red marble is the light one. And if the blue one is heavier, then the first red marble is the heavier one as well.

(iii) If the tray with the red and green marble is lighter, then the same process from (ii) would be followed but with the roles of the red and blue marbles reversed.

That covers all the possible scenarios, and in each one, only two weighings were required. I'm impressed that you have found three distinct ways to do it; it took me longer than expected to figure out just one way. Sorry it's a bit convoluted, but I just typed as I thought about it so it didn't end up being that concise. :(

Brian Charlesworth - 2 years, 8 months ago

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Great. I guess on analysis you could say the other two maybe termed as variations Brian. Posting one of them...

A)The first step is the same R1G1 on one side and G2B1 on the other side.

a) if equal just weigh R1 with B1

b) If unequal, say R1G1 is heavier, it means

1) Both R1G1 are heavy, both G2B1 are light

2) Both R1G1 are heavy, Only B1 is heavy

3) R1 is light, G1 heavy and both G2 B1 light.

So R1B1 can be both both heavy or both light or R1 is heavy and B1 is light. G1 is established as heavy.

B) Now measure R1B2 v/s G1B1

1) if equal, both R1B1 are light

2)If R1B2 is heavier, means R1 heavy and B1 light

3) R1B2 is lighter then both R1 B1 are heavy.

Satyen Nabar - 2 years, 8 months ago

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Ah, o.k.. and if R1G1 is lighter to start with, then step B) would involve different marbles, but the same idea. Fun question; I'm thinking now of a generalization of the problem where we have \(n\) pairs of similar marbles of \(n\) different colors. We could use a divide and conquer approach, i.e., take three pairs at a time and sort as before, and deal with any leftovers mod 3 separately. So for, say, \(9\) pairs of marbles we would require \(6\) weighings. However, there may be a more efficient approach using groups of more than three pairs at a time.

Brian Charlesworth - 2 years, 8 months ago

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