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Pumpkins 2 (Posted on 2005-09-08) Difficulty: 3 of 5
Five pumpkins are weighed two at a time in all possible combinations, similar to the first pumpkins puzzle. The results of the weighings gives nine different values: 52, 56, 60, 68, 72, 76, 80, 84, and 88. One of the values was repeated, but which value was not written down.
Find out the weights of the pumpkins and which value is the repeat.

See The Solution Submitted by Brian Smith    
Rating: 3.0000 (2 votes)

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Solution Solution | Comment 3 of 5 |
 
Since there is only one repeat weighing,
all the weights must be distinct.
Since all weighings are even, the weights
must be all odd or all even.
Let the weights be A < B < C < D < E.
A+B = 52 and A+C = 56 ==> C = B+4
C+E = 84 and D+E = 88 ==> D = C+4 = B+8
A+84 = A+(C+E) = (A+C)+E = 56+E ==> E = A+28
A < B < B+8 = D < E = A+28 ==> A < B < A+20
Therefore, A+B < 2B < (A+B)+20 
            52 < 2B < 72
            26 < B < 36
Hence, we have the following nine possibilities, 
    A   B   C   D   E 
1)  25  27  31  35  53     31+35 = 66
2)  24  28  32  36  52     28+36 = 64
3)  23  29  33  37  51     29+37 = 66
4)  22  30  34  38  50     30+34 = 64
5)  21  31  35  39  49     31+35 = 66
6)  20  32  36  40  48
7)  19  33  37  41  47     19+47 = 66
8)  18  34  38  42  46     18+46 = 64
9)  17  35  39  43  45     17+45 = 62
Only 6) satisfies the problem,
     +  32  36  40  48  
    20  52  56  60  68  
    32      68  72  80
    36          76  84
    40              88
As can be seen from the + table, 68 is the repeat.
Note: Possibilities 2), 4), and 8) could have been
      picked to define similar problems with
      different weighings and single repeats.
 

  Posted by Bractals on 2005-09-08 20:16:14
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