Much like the

third Pumpkins puzzle six pumpkins, each having a different weight, are weighed two at a time in all 15 sets of two. This time exactly four pairs of duplicate values occurred. Those four values are 60, 100, 110, and 120 pounds.

How much did each individual pumpkin weigh?

(In reply to

re(4): Analytical solution (Bonus puzzle question) by Brian Smith)

I'm still working.

I confirm that there are no solutions for {100,110,116,120} where a+d = b+c and c+f = d+e.

I do have solutions to other problems where b+c = 100 and c+f = d+e = 220.

(38,48,52,58,62,68) gives {100,106,110,120}

(39,47,53,59,61,67) gives {100,106,114,120}

The complementary solutions are obtained by subtracting each pumpkin from 110 and their totals from 220:

(42,48,52,58,62,72) gives {100,110,114,120}

(43,49,51,57,63,61) gives {100,106,112,120}

None of these are Brian's bonus problem. And, even if they were, I am not satisfied as I would not have demonstrated uniqueness.

I need a better analytical hammer.