"Family math": that's what my 6th-grader's math teacher calls these somewhat elaborate word problems that she hands out once per month. At first, I was alarmed, because: is my 6th-grader really supposed to be able to figure this out? But now that I know it's called "family math," I've formed the theory that the teacher knows these problems are over challenging for kids this age, and the idea, perhaps, is just to get them talking with a parent about a math problem. If so, here is the way it works at this house: "Dad, the family math problem is due Friday. Have you figured it out yet?"

Here's the problem due on Friday:

*You have five bales of hay. For some reason, instead of being weighed individually, they were weighed in all possible combinations of two: bales 1 and 2, bales 1 and 3, bales 1 and 4, bales 1 and 5, bales 2 and 3, bales 2 and 4, and so on. The weights of each of these combinations were written down and arranged in numerical order, without keeping track of which weight matched which pair of bales. The weights in kilograms were 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91.*

*Your initial task is to find out how much each bale weighs. In particular, you should determine if there is more than one possible set of weights, and explain how you know. Once you are done looking for solutions, look back over the problem to see if you can find some easier or more efficient way to find the weights.*

First observation would be that it's good there are ten data points, because there are indeed ten ways of "choosing two from five." If we complete the job of setting them out, using letters instead of numbers (so that the bale identifiers don't get mixed up with their weights), they are a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e, and d+e.

Second observation is that the heaviest bale has to weigh at least 46 kilograms (else there is no way for the heaviest result of 91 kilograms) and the lightest bale can be no more than 40 kilograms (else there is no way for the lightest result of 80 kilograms).

Third observation is that the ten data points range from a low of 80 to a high of 91—not much variation, indicating that there isn't very much variation in the weights of the individual bales, either.

Fourth observation is that no result occurs twice but two possible results within the range do not occur at all—namely, 81 and 89. I think this must mean that the weights of the individual bales, though tightly bunched, are nonconsecutive. (Notice, for example, that if you are making all possible simple sums of 1, 2, 3, and 4, the results are 3, 4, 5, 5, 6, and 7—two 5s and no "gaps." We have gaps but no double results.) Also, if two individual bales had the same weight, then there would have to be a "double result"—for example, if a and b are both 40, then a+c=b+c, and you'd have the same result twice.

Last observation is that one of our gaps (81) occurs near the low end of the range of data points and the other (89) occurs toward the high end of the range. Since the gaps in the outputs occur near the low end and near the high end of the range, the gaps in the inputs are likely also toward the bookends.

If we put this all together, it seems a reasonable guess might be that the two lightest bales weigh 39 and 41 kilograms (thus yielding the 80 result) and that the two heaviest weigh 44 and 47 kilograms (thus yielding the 91 result). Let's see if we can make that work.

If our individual bales a through e are in ascending order of weight, then we are guessing that a is 39, b is 41, d is 44, and e is 47. So,

a+b=39+41=80

a+c=39+c

a+d=39+44=83

a+e=39+47=86

b+c=41+c

b+d=41+44=85

b+e=41+47=88

c+d=c+44

c+e=c+47

d+e=91

The six completed sums—80, 83, 85, 86, 88, and 91—match up with six of the results that we're given. The four incomplete sums are 39+c, 41+c, 44+c, and 47+c. We're hoping to match them to the four results that haven't been accounted for, which are 82, 84, 87, and 90. And they can be matched if c is 43. One solution, therefore, is that the five bales weigh 39, 41, 43, 44, and 47 kilograms.

When looking for more solutions, it seems there's a pretty small number of possibilities that don't contradict any of the above observations. For example, if a is 36 and b is 44 (thus yielding the lightest result of 80), the range becomes too wide and there is no way to hold the heaviest result to just 91. If a is 37 and b is 43, the others could be 44, 45, and 46—but you can't have both 44 and 45, since 89 is not among the results. You have to run through the possibilities related to a being 38 and b being 42, but none of them can be made to work. I think the solution we've found must be the only one.

Not going to be easy making Lydia take an interest in this.

UPDATE: She was less resistant than I guessed and, in talking through it with her, I thought of a more "elegant" method. It seems almost trivial to note that a+b has to be 80, and d+e has to be 91. An "observation" that I should have made, as it's at least as helpful as any I mentioned, is that a+c has to be the second lightest result (82) and c+e has to be the second heaviest result (90). The possible weights for bales (a, b), then, are (40, 40), (39, 41), (38, 42), (37, 43), etc. But (40, 40) doesn't work, since if a=b then, for example, a+c=b+c, and it doesn't, since there are no "double" results—each result is unique. If a is 39, then b is 41, and c is 43, so that a+c=82. Further, if c is 43, then e is 47, so that c+e=90; and in that case, d has to be 44 in order to make d+e=91. That's the solution we arrived at above.

Next check (38, 42) for (a, b). Then c has to be 44, d has to 45, and e has to be 46. But in that case, c+d=89, and 89 is not one of the results.

Next check (37, 43) for (a, b). Then, for a+c to be 82, c has to be 45. But now there is no way to hold the heaviest result to only 91. The same problem arises for (36, 44), (35, 45), (34, 46), etc. So the only solution is a=39, b=41, c=43, d=44, and e=47.

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