You have a scale which compares two weights but instead of telling which is heavier or lighter, this scale returns the difference of the weights but will not tell which is heavier. In order for the scale to display an accurate result, the amount of weight on each side must be at least 20g and the difference can not exceed 20g. The scale will not report a result unless there is a weight on each side. If the difference exceeds 20g, the scale will report an error.

Use this scale no more than six times, sort an otherwise identical set of six coins with weights of 20, 21, 22, 23, 24, and 26 grams.

Examples:
Ex 1: If A=30 B=30 then the scale reports 0 (equal)
Ex 2: If A=31 B=29 then the scale reports 2
Ex 3: If A=29 B=31 then the scale reports 2 (undistinguishable from example 2)

Randmly pick one of the six coins and weigh it against remaining coins. Following table sums up the possible results:

Random coin::   20g    21g    22g   23g    24g   26g |Wt Code  
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      20               X       1       2       3       4       6     (12346)       
      21               1       X       1       2       3       5     (11235)   
      22               2       1       X       1       2       4     (11224)
      23               3       2       1       X       1       3     (11233)
      24               4       3       2       1       X       2     (12234)
      26               6       5       4       3       2       X     (23456)

Different Wt code for all the six cases indicates randomly selected coin can be identified after above step.

For different wt codes sixth weighing should be as under:
1) 12346 :- Problem already solved.
2) 11235 :- Pick one of the coins which earlier gave difference 1 (say coin A) and weigh it against coin with diff 2 (i.e 23g). Now weight of coin A would be 22g or 20g for the result of the 6th weight to be 1 or 3 respectively. Problem Solved.
3) 11224 :- Put 22g (earlier randomly selected coin)and 26g(coin that gave diff of 4) on one side. On the second side we would put a pair of 2 coins which gave a difference of 1 and 2. Possible combinations are 20+21, 21+24, 20+23 and 23+24. For these combinations outcome of sixth weigh will be 7,3,5 or 1. Since all are unique, all coins now can be uniquely detemined.
4) 11233 :- We need to repeat above step with 21g and 23g on one side. And pair of coins which earlier gave a difference of 1 & 3. Same as above I found unique outcome of sixth weigh. Problem solved again.
5) 12234 :- Now this has become intuitive and repetitive. It is solved same as done for Wt code 11235.
6) 23456 :- Already solved.

That sums it all.

Can we do it by weighing 5 times only ??

Posted by Rupesh Khandelwal on 2005-07-04 11:30:14