Let A-F be the weights of the pumpkins in ascending order.
There are 15 possible ways for a duplicate weighing to occur:
A+D = B+C, A+E = B+C, A+E = B+D, A+E = C+D, B+E = C+D,
A+F = B+C, A+F = B+D, A+F = C+D, B+F = C+D, A+F = B+E,
A+F = C+E, B+F = C+E, A+F = D+E, B+F = D+E, C+F = D+E.
Finding valid sets of four equalities is made easier as the 15 equalities can be divided into six subsets such that each subset can only contribute only one (or in one case two) members to a potential solution:
Subset 1: A+D = B+C
Subset 2: A+E = B+C, A+E = B+D, A+E = C+D
Subset 3: A+F = B+C, A+F = B+D, A+F = C+D, A+F = B+E, A+F = C+E, A+F = D+E
Subset 4: B+E = C+D
Subset 5: B+F = C+D, B+F = C+E, B+F = D+E
Subset 6: C+F = D+E
The one case where a subset can contribute two members is in subset 3 with A+F = B+E and A+F = C+D.
Certain pairs, such as A+D = B+C vs A+E = B+C, can be eliminated as part of a solution immediately. That is because they would imply two of the pumpkins are equal, which is not the case.
There are still several ways to find suitable groups of four equalities, but they can be classified in four different ways:
Case 1: The equalities imply all six pumpkins are in arithmetic progression. Example: A+D = B+C, A+E = B+D, B+E = C+D, B+F = C+E
This creates five different duplicate sums, so cannot be a basis to form a solution.
Case 2: The equalities imply five pumpkins are in arithmetic progression, plus one hanging on. Example: A+D = B+C, A+E = B+D, B+E = C+D, A+F = D+E.
Let the arithmetic progression be x, x+d, x+2d, x+3d, and x+4d; let the sixth pumpkin be y.
Then three of the duplicate weights are x+3d, x+4d, and x+5d. The three weighings 100, 110, and 120 imply the arithmetic progression is 35, 45, 55, 65, 75.
The possible values for the fourth duplicate weighing from this set are 80, 90, 130, and 140. But 60 is not one of those values so no value of y can exist to make this a solution. (If 80 was the fourth weighing then y=5 would create a solution for a problem asking for 80, 100, 110, 120 as the duplicates.)
Case 3: The equalities imply four pumpkins are in arithmetic progression, plus the other two pumpkins have the same difference. Example: A+F = B+E, A+E = B+D, A+D = B+C, C+F = D+E
Let the arithmetic progression be x, x+d, x+2d, and x+3d; let the other two pumpkins be y and y+d.
The four duplicate weights are x+3d, x+y+d, x+y+2d, and x+y+3d. The last three must equal 100, 110 and 120; the first equals 60.
This yields a solution 15, 25, 35, 45, 75, 85.
Case 4: The equalities imply two arithmetic progressions of three terms. Example: A+D = B+C, C+F = D+E, A+F = B+E, A+F = C+D
Let the arithmetic progressions be x, x+d, and x+2d; y, y+d, and y+2d.
This yields only three distinct weights since there is a triplicate weighing in the group [x+(y+2d) = (x+d)+(y+d) = (x+2d)+y].
A fourth duplicate can be made if x+(x+2d) = y+(y+d), but this implies x+d/2=y, which makes all six weights into a single arithmetic progression like Case 1, so there can be no solution from this case.
In summary, the unique solution is the pumpkins' weights are 15, 25, 35, 45, 75, and 85 pounds.
The bonus puzzle in the comments asking about weights 100, 110, 116, 120 can be solved the same way yielding pumpkins weighing 43, 47, 53, 57, 63, and 73 pounds. |
