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 Two sets of Coins (Posted on 2005-10-07)
You have two sets of five coins. Each set has four real 30g coins and one 31g fake coin. It is easy to find the fake coins using a balance scale a total of four times, two weighings for each set.

Can you find a strategy which uses only three weighings?

 See The Solution Submitted by Brian Smith Rating: 3.5000 (2 votes)

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 re: not sure i get your logic for scenario two... | Comment 5 of 7 |

You made this an independent comment, so I'm not sure it's in response to my first comment (comment # 2 altogether for this puzzle), but I assume it is. I do see there's a mistype on my comment: I list Ed twice-- one should be Ee.

Ady, in comment #1 had said if ABC = abc, he'd next weigh AB vs de, not the D vs d you specify.  I don't know if you are suggesting an improvement over Ady's method.

In any case, Suppose you found ABC = abc and then went on to do D vs d and E vs e. There are 9 possible outcomes, which I show here with the implications for the identities of the two heavy weights:

D<d and E<e: can't happen as one capital and one lower case must be heavy.

D<d and E=e: can't happen as then one of A,B or C would be heavy and first weighing wouldn't have come out the way it did.

D<d and E>e: E and d are the heavies

D=d and E<e: can't happen as then one of A,B or C would be heavy and first weighing wouldn't have come out the way it did.

D=d and E=e: Either D and d are the heavies or E and e or A and a or A and b or A and c or B and a or B and b or B and c or C and a or C and b or C and c; you haven't resolved which, and you've already used your 3 weighings.

D=d and E>e: can't happen as then one of a,b or c would be heavy and first weighing wouldn't have come out the way it did.

D>d and E<e: D and e are the heavies.

D>d and E=e: can't happen as then one of a, b or c would be heavy and first weighing wouldn't have come out the way it did.

D>d and E>e: can't happen as then one of a, b or c would be heavy and first weighing wouldn't have come out the way it did.

And even if a weighing strategy were devised where the second weighing depended on the first, there'd still be only 9 possible outcomes for the sequence of two (dependent) weighings: left-left; left-equal; left-right; equal-left; equal-equal; equal-right; right-left; right-equal; right-right; but there are 13 possibilities to distinguish, each requiring a unique set of weighings to be identified: Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb, Cc, Dd, De, Ed, or Ee.

 Posted by Charlie on 2005-10-21 14:01:06

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