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?
Let the pumpkins weigh a,b,c,d,e,f in order of increasing weight.
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
2) Similarly, 120 must = c+f = d+e.
So, f = 120-c = 60+b, and e = 120-d = 60+a
The 6 pumpkins (in increasing order) weigh a,b,60-b,60-a,60+a,60+b
3) b<(60-b), so b < 30
4) At least 8 of the 15 weighings are divisible by 10. 8 is more than half of 15, so all of the pumpkins must have weights that are multiples of 5.
5) 60+a+b = a+f = b+e, so that is one of the duplicate weighings, and it must be either 100 or 110
5a) if 60+a+b = 110, then a+b = 50
But there are no values of a+b=50 such that a < b < 30 and a and b are multiples of 5.
5b) Therefore, 60+a+b = 100, so a+b = 40
There is only one possiblity of a+b=40 such that a < b < 30 and a and b are multiples of 5.
Namely a = 15, b = 25
THEREFORE, the pumpkins can only be (15,25,35,45,75,85).
Checking: 110 = 25+85 = 35+75, so everything works
Analytical solution
