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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.

  Submitted by Brian Smith    
Rating: 3.0000 (2 votes)
Solution: (Hide)
My solution is below. Bractals presents a solution here.

Let A>B>C>D>E and let w represent the repeated weight. The following equations can be formulated:
A+B=88 - sum of the two heaviest weights
A+C=84 - sum of the heaviest and middle weights
C+E=56 - sum of the lightest and middle weights
D+E=52 - sum of the two lightest
4*(A+B+C+D+E) - w = 636 - the sum of all the weighings

w can be any of 60, 68, 72, 76 or 80. Substituting those values for w yields five different sets for A,B,C,D, and E. Each set of values is summed in pairs to see if that set gives the same nine values as the original values.

 w | A | B | C | D | E |
---+---+---+---+---+---+-----
 60| 50| 38| 34| 30| 22| No pair sums to 76
 68| 48| 40| 36| 32| 20| Solution
 72| 47| 41| 37| 33| 19| No pair sums to 68
 76| 46| 42| 38| 34| 18| No pair sums to 68
 80| 45| 43| 39| 35| 17| No pair sums to 68
Since 48, 40, 36, 32, 20 is the only set of values which sums properly, it must be the solution.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionPuzzle SolutionK Sengupta2008-11-12 03:47:05
AnswerK Sengupta2008-11-10 08:46:18
SolutionSolutionBractals2005-09-08 20:16:14
Solutionthe Computer wayCharlie2005-09-08 20:08:23
SolutionSolutionAndre2005-09-08 19:13:58
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