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The weights of the bales were 39, 41, 43, 44, and 47. Here are the calculations:
80 = 39 + 41 (1 & 2)
82 = 39 + 43 (1 & 3)
83 = 39 + 44 (1 & 4)
84 = 41 + 43 (2 & 3)
85 = 41 + 44 (2 & 4)
86 = 39 + 47 (1 & 5)
87 = 43 + 44 (3 & 4)
88 = 41 + 47 (2 & 5)
90 = 43 + 47 (3 & 5)
91 = 44 + 47 (4 & 5)
We could not find any other set of weights and we believe that this is the only set of possible weights...
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I think this is a good P.O.W. because it teaches you how to solve things, but I personally think in reality, the hay baler could learn to not be lazy and just reweigh the bales. From this problem, I learned to problem solve and keep experiment through trial and error until you get your answer. I think the problem could have given us the weight of one bales so we could use an algebraic expression to figure out the weights of the other bales...
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