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 Analytical solution
by Steve Herman)
Nice work, but how can you be sure that your statement "1) a+b and a+c are both less than all other pairs, so 60 must = a+d = b+c. Therefore, d = 60-a and c = 60-b" is true. It could have been the case that a+e=60 with one of b+c, b+c, or c+d being the other pair.