Weighing Bales of Hay

Date: 12/10/97 at 19:34:17
From: Anthony D Mays
Subject: POW(Problem of the Week)

Dear Dr. Math,

Here's the problem.

You have five bales of hay. For some reason, instead of being weighed
individually, they have been 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 weight of each of these combinations has been written down and 
they have been arranged in numerical order, without keeping track of 
which weight matches which pair of bales. The weights in kilograms are 
80, 82, 83, 84, 85, 86, 87, 88, 90, 91. 

How much does each bale weigh? Is there more than one possible set of 
solutions? Explain your answer.

I attempted to start, but all I could do was list all the possible
combinations that I could find. I believe, though, that the weights 
are between 30 and 40 kilograms.

Date: 12/11/97 at 00:40:01
From: Doctor Pete
Subject: Re: POW(Problem of the Week)

Hi,

First of all, call the weights of bales 1 through 5 a, b, c, d, e.  
Since the ten weighings gave different results, each bale must have a 
distinct weight from the others (why?).  So assume a < b < c < d < e.  
Then

     a + b < a + c < b + c

and similarly,

     d + e > c + e > c + d

Hence a + b = 80 is the lightest pair, a + c = 82 the second lightest, 
d + e = 91 the heaviest, and c + e = 90, the second heaviest.  

Notice that we cannot say b + c = 83, because we do not yet know if 
a + d < b + c.  But from what we do know,

     a+b = 80       d+e = 91
     a+c = 82       c+e = 90
     --------       --------
     c-b = 2        d-c = 1 

Hence c = b+2, and d = c+1 = b+3. Hence a+d = a+b+3 = 83, so in 
particular we have

     a < 80-a < 82-a < 83-a < a+8

From this we find that b+e = 88. But 83 is the third lightest pair, 
and 88 the third heaviest. Therefore, a+d < b+c, and b+e > c+d, and 
we must conclude that b+c is the fourth lightest, and c+d is the 
fourth heaviest; i.e.,

     b+c = 84,  c+d = 87.
So
     a+b < a+c < a+d < b+c <     <     < c+d < b+e < c+e < d+e
      80    82    83    84    85    86    87    88    90    91 

We don't need to fill in the rest, because we immediately find that

     b+c = (80-a) + (82-a) = 162 - 2a = 84

which implies a = 39.  Finally,

     a = 39, b = 41, c = 43, d = 44, e = 47.

Adding them gives the above, with b+d = 85, and a+e = 86.  The 
solution, by our analysis, is unique.

Comments: There are several important lessons to be learned from this 
problem. First, the observation that the paired weights are distinct 
tells you the individual weights must also be distinct, and therefore 
must have an ordering. The ordering and assignment of paired weights 
leads to a system of equations that can be solved, but not always.  
(Why?)

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/