This is in continuation of
Pumpkins and
Pumpkins 2.
Six pumpkins - each having a different weight - are weighed two at a time in all 15 sets of two. The weights are recorded as: 27, 36, 39, 45, 48, 51, 54, 57, 63, 66, 72, 75 and 81 pounds.
- Precisely two of the values occurred exactly twice, but these values were only written down once.
- Each of the individual weights (in pounds) is a positive integer.
Derive the weights of the pumpkins and which two values occur twice.
Let A-F represent the pumpkins in ascending order of weight. Then A+B = 27, A+C = 36, D+F = 75, and E+F = 81.
No matter what the equal pairs are, the following are always true:
A+B < A+C < A+D < A+E < {A+F and B+E} < B+F < C+F < D+F < E+F
(A+B)+(E+F) = (A+E)+(B+F) = (A+F)+(B+E) = 27+81 = 108.
There are three other pairs of weighings that can add to 108 and be applicable to A+E, A+F, B+E, and B+F. Those are 45+63, 51+57, and 54+54.
This leads to four possible scenarios to combine with the four equations at the beginning:
Case 1: A+E = 45, B+F = 63, A+F = 51, B+E = 75
Case 2: A+E = 45, B+F = 63, A+F = 75, B+E = 51
Case 3: A+E = 45, B+F = 63, A+F = 54, B+E = 54
Case 4: A+E = 51, B+F = 57, A+F = 54, B+E = 54
Case 1 leads to C=28.5 and D=31.5, which add up to 60. Not a solution.
Case 2 leads to C=25.5 and E=34.5, which add up to 60. Not a solution.
Case 3 leads to A=9, B=18, C=27, D=30, E=36, and F=45. But no sum adds to 51, so is not a solution.
Case 4 leads to A=12, B=15, C=24, D=33, E=39, and F=42, with duplicate weighings of 54 and 57.
(Fixed an oversight Steve spotted.)
Edited on January 26, 2016, 11:13 am
Another Analytic Solution
