Weighing Bales of Hay

Date: 2/2/96 at 2:22:3
From: Anonymous
Subject: The Haybaler Problem

You have 5 bales of hay. Instead of being weighed individually, they 
were weighed in all possible combinations of two: bales 1 and 2, 
1 and 3, 1 and 4, 1 and 5, bales 2 and 3, and bales 2 and  4 etc. 
The weights of each of these combinations were written down and  
arranged in numeric order, without keeping track of which weight 
matched which pair of bales. The weights in kilograms are 110, 112, 
113, 114, 115, 116, 117, 118, 120, and 121. 

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

Date: 6/17/96 at 14:36:1
From: Doctor Alain
Subject: Re: The Haybaler Problem

Let the 5 bales be in increasing weight order and call their weights 
B1, B2, B3, B4 and B5. There can't be 2 bales of the same weight or 
else there would be pairs of bales with the same weight (if B1 = B2 
then B1 + B3 = B2 + B3). The lightest pair is 110 kg so

(i) B1 + B2 = 110.

The second lightest pair is 112 kg so

(ii) B1 + B3 = 112.

The heaviest pair is 121 kg so

(iii) B4 + B5 = 121.

The second heaviest pair is 120 kg so

(iv) B3 + B5 = 120.

From (iv) and (ii) we get B5 = B1 + 8. If we put this back in (iii)
we get 

(v) B1 + B4 = 113.

so the third lightest pair is bale 1 and bale 4. The fourth lightest pair 
can be bails 1 and 5 OR bales 2 and 3. 

So (vi.a) B1 + B5 = 114    OR    (vi.b) B2 + B3 = 114.

Try each of these equations. One works out fine - the other doesn't.

-Doctor Alain,  The Math Forum