This is in continuation of
Pumpkins 3.
Six pumpkins - each having a different weight - are weighed two at a time in all 15 sets of two. The weights are recorded as: 30, 33, 39, 45, 48, 51, 54, 57, 60, 63, 69, 75 and 78 pounds.
- Precisely one of the values occurred exactly three times, but that value was only written down once.
Derive the weights of the pumpkins and which value occur thrice.
Let the six pumpkins, smallest to largest, be a, b, c, d, e and f
Necessarily
30 = a + b
33 = a + c
78 = e + f
75 = d + f
Well, having learned from "Pumpkins 3", let's try the easy solution first. One possible solution is:
39 = b + c
69 = e + d
This leads simply and directly to:
[a,b,c,d,e,f} = {12,18,21,33,36,42}
This proves to be a solution, as the pairwise sums are
30, 33, 39, 45, 48, 51, 54, 54, 54, 57, 60, 63, 69, 75, and 78
However, this is not necessarily the only solution.
It is possible that
39 = a + d
69 = c + f
This requires more work.
A full solution will follow.
First solution (partial spoiler?)
