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: Analytical solution
by Brian Smith)
Good point, Brian. Yes, I did not prove that a + e <> 60. It just did not seem possible to me that a + e = 60, and in fact it is not possible.
Here is a belated proof.
If a + e = 60, then e < 60, so a and b and c and d are also < 60. Then there is no way to have two different pairs sum up to 120. So a+e <> 60.
Edited on January 23, 2016, 12:11 pm