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(6): Analytical solution (Bonus puzzle question) by Steve Herman)

Thank you Steve. Actually creating this problem took a while itself.

I referenced Pumpkins 3 in the problem statement because it was while solving that problem that I came formulated this problem. I was actually trying to create a solution like I had for Pumpkins 4, which leveraged the triple equality A+F=B+E=C+D. But I could not find any way to decisively split the weights unless there were more equal pairs. I eventually realized if I had 4 equal weighings that those weighings alone could uniquely determine the individual pumpkins weights. Thus this puzzle was born.