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Twelve Coins and a Broken Scale (Posted on 2005-06-24) Difficulty: 4 of 5
You have 12 coins, six weigh 24 grams and six weigh 25 grams. You also have the broken scale from Five Weights and a Broken Scale.

Sort the 12 coins into the group of 24g coins and the group of 25g coins using that broken scale no more than 9 times.

  Submitted by Brian Smith    
Rating: 3.3333 (3 votes)
Solution: (Hide)
At most nine weighings are needed as follows.

Note: I will refer to the 25g coins as heavy coins and 24g coins as light coins.

Divide the coins into four groups (A,B,C,D) of 3 coins. Take three weighings: AB vs CD, AC vs BD, and AD vs BC By taking 6 vs 6 for the initial weighings, if two groups are different, there is enough of a difference(2g+) for the broken scale to sense it.

Case 1: None of the weighings are equal
One of the groups is always on the heavy side or always on the light side of the weighings
For example, if the weighings were AB>CD, AC<BD, ADCD, AC>BD, AD<BC then D is always light, goto Case 1.2 Case 1.1:One group of three is always heavy
All three coins in the heavy group are known to be heavy coins. Each of the other groups has exactly 1 heavy and 2 light coins.
Take one coin from the heavy group and call it K. Choose any one of the mixed groups and label its three coins X, Y and Z. Make two weighings: KX vs YZ and KY vs XZ
If the results are KX=YZ and KY=XZ then Z is heavy and X and Y are light. If the results are KX=YZ and KY>XZ then Y is heavy and X and Z are light. If the results are KX>YZ and KY=XZ then X is heavy and Y and Z are light.
Repeat the procedure for the other two mixed groups.
Total weighings: 9

Case 1.2:One group of three is always light
This is identical to Case 1.1 with heavy and light switched around.
Also total weighings: 9

Case 2: Exactly one weighing is equal
One group is three heavy coins, one group is three light coins, one group is two heavy and one light coin and one group is one heavy and two light coins.
If the weighings were AB>CD, AC<BD, and AD=BC then D is three heavy coins, C is three light coins and A and B are the two mixed groups. Make three weighings: (X1,X2,Y3)vs(Y1,Y2,X3), (X1,Y2,Y3)vs(Y1,X2,X3), and (X1,Y2,X3)vs(Y1,X2,Y3).
Two of the weighings will be equal, the one other weighing will not be equal. The three coins on the heavy side of that weighing are all heavy and the three coins on the light side are all light.
Total weighings: 6

Case 3: Two weighings are equal, one is not equal.
Choose two groups on different sides of the unequal weighing. Label the coins from one group X1, X2 and X3. Label the coins from the other group Y1, Y2 and Y3.
Make three weighings: (X1,X2,Y3)vs(Y1,Y2,X3), (X1,Y2,Y3)vs(Y1,X2,X3), and (X1,Y2,X3)vs(Y1,X2,Y3).

Case 3.1:All three weighings are equal
Both groups on the heavy side of the original unequal weighing consist of all the heavy coins and both groups on the light side consist of all the light coins.
Total weighings: 6

Case 3.2:There is one unequal weighing
The three coins on the heavy side of that weighing are all heavy and the three coins on the light side are all light. That established six coins. Take the other two groups and label and weigh them the same way to finish.
Total weighings: 9

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionFixed solution...Cougar Draven2005-07-05 22:18:19
SolutionSolution (by Visual Basic program)Penny2005-07-03 19:19:24
SolutionSolution (By Ed Mekler and Chris LongfellowChris Longfellow2005-07-02 22:18:43
Solution =)Eugene2005-06-30 06:34:55
Some ThoughtsSome thoughtsowl2005-06-30 04:06:21
re(2): Solution, I think.Penny2005-06-28 14:13:53
re: Solution, I think.Federico Kereki2005-06-26 01:34:23
SolutionSolution, I think.Cougar Draven2005-06-25 22:44:59
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