Mary has an unusual ring. It is made up of a set of three interlocking ringlets, each set with a large gemstone, a diamond, a ruby, and a sapphire. It is very valuable and the insurance company insists it be kept locked in their vault, except on those occassions where Mary is wearing it. Because of this, Mary had a copy made, which she wears on lesser occassions.

Today she attended an affair in which she was able to convince the insurance company that she needed to wear the original. Half an hour after returning home the insurance rep called to let her know he was on his way to pick up the ring.

Mary then realized that she'd absent-mindedly taken off the ring and put it in her jewelry box, where she keeps the copy.

When it's not on her finger, the ring separates into its componants, so Mary was looking at six nearly identical ringlets, two with blue stones, two with red, and two with white.

She needs to separate the genuine ringlets from the copies. She knows that each of the ringlets in each set weighs the same.(That is the genuine ringlets each weigh the same, and the copies each weigh the same.) She also knows that a copy weighs less than its original.

The only scale she has that is delicate enough to properly weigh the ringlets is a small balance scale she uses to measure headache powders and sleeping draughts (she can't swallow pills), but the weights tha she uses for her medicine are of an order too small for the ringlets. She will have to weigh them against each other. She could do it in three weighings by trying each against its counterpart, but she is certain to be "caught" before she finishes, and either her insurance will go up, or the company will be "forced" to not allow her to wear the ring any more.

Is there a way to separate the six ringlets in two weighings?

Maybe like this:

Separate the ringlets in two groups (just by colour):
Group 1: B1, R1, W1
Group 2: B2, R2, W2

First weighing:
- B1, R1 on the left side
- W2, R2 on the right side

Note that R1 and R2 are of different weights.

There are three possibilities:
1. The left side goes down. In this case R1 can't be lighter than R2, otherwise the scale would have balanced or the right side would have gone down. Therefore, we can conclude that R1 is original and R2 is copy. However, we don't know if B1 and W2 are both heavy or light or if B1 is heavier than W2 (W2 can't be heavier than B1 because the scale would have balanced). Therefore, we go to the second weighing:
- R1, R2 on the left side
- B1, W2 on the right side
Again three possiblities:
a) The left side goes down. This means that, since R1 is original and R2 is copy, both B1 and W2 are copies (lighter). Therefore, B2, R1 and W1 are originals;
b) The right side goes down. Both B1 and W2 are originals, making a set with R1;
c) The scale balances. B1 is heavier than W2 and therefore B1, R1 and W1 are originals.

2. The right side goes down. Analogous to the possibility 1. We conclude that R2 is original and R1 copy. We proceed with the second weighing analogous to the possibility 1).

3. The scale balances. This means that R1 and W2 belong together, and B1 and R2 belong together. We still don't know which set is lighter. Therefore we go to the second weighing:
- B2 on the left side
- W1 on the right side
If B2 goes down, then B2, R1 and W2 are originals, and if W1 goes down, then B1, R2 and W1 are originals.

Of course, maybe there's a simpler solution than this....

Posted by lucky on 2002-09-20 12:04:22