At Penn Alexander’s math club yesterday, the students worked on a fun puzzle that I’d never seen before. It goes like this:
You have five bales of hay.
For some reason, instead of being weighed individually, they were weighed in all possible combinations of two. 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.
How much does each bale weigh? Is there a solution? Are there multiple possible solutions?
Unfortunately, the problem seemed a little beyond them (or at least, they thought it was beyond them, so they quickly lost interest) but this seems like a great problem to use in middle school or high school math classes. In middle school, keep them talking and focus on the methods they employ to try to solve it. In high school, perhaps once they solve it you could get them to try generalizing the problem (to other sets of weights, more than five bales, etc.).
The Math Less Traveled 
Let the weights be a, b, c, d, e in increasing order
e+d = 91
e+c = 90
So, d= c+1
a+b = 80
a+c = 82
=> b= c-2
So, 82-c<c-2<c<c+1<90-c are a<b<c<d<e
So, the possible sums are
80, 82, 83, 172-2c, 2c-2, 2c+1, 2c-1, 88, 90, 91
So, 84, 85, 86, 87 are 2c-1, 2c+1, 2c-2, 172-2c in some order
So, 2c-1, 2c+1 are 85, 87
So, c = 43
So, the five weights are 39, 41, 43, 44, 47 🙂
so you got c = 43 and d = 44 => c + d = 87. But the orig. equation is c + d = 88 ??
Elegant! I had a solution, and felt confident it was the only one, but wasn’t sure how I’d go about proving it with the process I used. Getting down to one variable made your solution much nicer.
Nice. Much more elegant than my solution, and more generalizable, to boot. I did the following:
I noted that four of the combined weights were odd, which means that either all but one of the individual weights were even, or all but one were odd. Call the misfit a, and the remaining individual weights b, c, d, and e, in increasing order (that is, b<c<d<e).
Thus we have
83 = a+b ⇒ b = 83-a
85 = a+c ⇒ c = 85-a
87 = a+d ⇒ d = 87-a
91 = a+e ⇒ e = 91-a.
We note that the smallest combined weight must be the sum of the smallest two individual weights. This gives us 80 = b+c. Substituting the values from above for b and c, we have
80 = (83-a)+(85-a) = 168 – 2a ⇒ 2a = 88 ⇒ a = 44.
Substituting a into the original four equations, we have
b = 39
c = 41
d = 43
e = 47.
Thus the original weights are 39, 41, 43, 44, and 47 kilograms.
xander
This is funny! I solved the problem. Then, upon reading the three comments above, I realized that I had read the problem wrong. For some reason I thought that the problem stated that the bales of hay were weighed in all possible combinations of two and then the AVERAGE of the weights were recorded as 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91.
It turns out that doing the problem this way just adds a couple of steps and the weights come out to twice the actual answer (78, 82, 86, 88, and 94 as compared to 39, 41, 43, 44 , and 47).
But since Brent’s original post suggested a way of making the problem more challenging for the High School classroom, I thought this might make an interesting suggestion (even if I did arrive at it accidentally).
I admit, I enjoy your not-so-subtle ongoing crusade against the tendency to assess a problem as difficult and give up without trying one’s best.
I started off like Pratik, writing equations from the lowest and highest pairs:
a+b = 80
a+c = 82
e+c = 90
e+d = 91
But for my fifth equation, I related (a+b+c+d+e) to the sum of all the given paired weights, in which each bale appears 4 times.
Well, I did it similarly to Craig T.Nelson there, working off the sum and the differences at the ends.
One of the reasons this one comes up so often is that there are multiple productive paths.
Here’s another: Each is weighed four times, so we can discover the sum of all weights by adding the list and dividing by four: 214.
As commenters noted: a+b=80. e+d=91. and 214 – 80 – 91 = 43. which is c.
A generalization of this is actually a classic problem, although I can’t find my original source. One reference that I could find was page 175 of Andreescu’s and Feng’s book “A path to combinatorics for undergraduates: counting strategies” (in the chapter on generating functions). Snippets of this book are available on the usual sources online.
The version with five bales is nice, and we used such a problem on the team round of the 2005 UGA Mathematical Tournament: http://www.math.uga.edu/olympiad/05/student-team05.pdf
The classic generalization is very simple: suppose there are n bales. Then, it is possible to (always) uniquely recover the weights if and only if n is not a power of two. For example, clearly one cannot recover the weights if n=1 (you have no data) or n=2 (you have only the sum of the weights). The case n=3 is good warmup for the n=5 case.
It’s a fun problem to think of counterexamples for n=2^k. Some kids might be able to get n=4, but n=8 is harder. The general solution is quite nice and some of this blog’s readers may like to work on it. (If you’re stuck, here’s a hint in rot13: Guhr-Zbefr frdhrapr.)
Proving that it is always possible when n is not a power of two is usually done with generating functions, as in the source by Andreescu and Feng above. NB: The solution can be shortened a bit by using l’HĂ´pital’s rule.
Enjoy!
Boris: thanks! I knew there must be generalizations and I was hoping someone would mention some in a comment. =)