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.
Just of interest, note that all pairwise sums are 3. Unlike "Pumpkins 3", however, pumpkin weights are not required to be integers. However, they all need to be multiples of 1.5. This turns out not to be an issue, per the following full solution:
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
We can also conclude that 39 = b + c or 39 = a + d, or both
Also, 69 = e + d or 69 = c + f, or both
CASE I)
If 39 = b + c and 69 = e + d, then 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
CASE II)
If 39 = b + c and 69 = c + f then this leads simply and directly to:
{a,b,c,d,e,f} = {12,18,21,27,30,48}
This does not work, because 12 + 30 = 42, which is not one of our sums
CASE III)
If 39 = a + d and 69 = e + d then this leads simply and directly to:
{a,b,c,d,e,f} = {6,24,27,33,36,42}
This does not work, in part because because 6 + 36 = 42, which is not one of our sums
CASE IV)
If 39 = a + d and 69 = c + f then this leads to:
{a,b,c,d,e,f} = {a, 30 - a, 33 - a, 39 - a, 42 - a, 36 + a}
where a can be any multiple of 1.5 between 4.5 and 13.5.
All possible solutions of this form fail, because
a + (42 - a) = 42, which is not one of our sums
So, Case I is the only solution.
Full Solution (spoiler)
