Contrary to Dr. Toom's reaction, I found the problem of the bales of hay to be intriguing (perhaps because HE was "bored" and did not find a solution in an hour, and because the wording of the write-up requirements suggests strongly that the problem has at least one solution).
I found that the write-up requirements encouraged me to do a bit more than just putter around with guess and check. In less than a half hour, I did the following. (Someone please tell me if I'm missing something here!)
I wrote down all the sums and found that the grand total was 856. I labeled the weights of the bales with the letters a through e. I mechanically wrote down the 10 combinations of 5 letters two at a time. Since their grand sum is 4a + 4b + 4c + 4d + 4e, I knew this sum must equal 856. So a + b + c + d + e = 214. Their average is 42.8.
So big deal. What does this mean?
Of the five numbers, either 2 are even or 4 are even.
Of the ten sums, 4 are odd. If 2 of the five numbers are odd, 6 of the sums would be odd, not 4. So there must be 4 odd and 1 even number.
It makes sense, therefore, to do the guessing and checking with lots of odd numbers centered around 43.
Soon I found this set: 39, 41, 43, 44, 47
I like the problem, though it's not "real world." It gives students an opportunity to examine explicitly some of their knowledge about odds and evens. It doesn't require the use of a calculator.
I am stopping work on it short of exploring the possibility of other solutions, which I think would be interesting--because my own "real world" demands my attention. I suspect that the solution is unique (since the sums are pretty tight), but invite others to continue.
It seems that my morning "arithmetic calisthenics" were just the thing Dr. Toom seems to want in our math classes. Perhaps my "testimony" will help him see the potential value in the problem.