Haybaler POW

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IAG 1H POW # 8: The Haybaler Problem

Word.PNGA Microsoft Word version of this work is available here: Image:Haybaler_Problem.doc

Pdf.jpgA PDF version of this work is available here: Image:Haybaler_Problem.pdf


Contents

Problem Statement

Not necessary to do.

Process

I started this POW with a lot of work which I didn’t need to do. I wrote down all the combos in the form of 1+2, 2+3, etc. I then guess some numbers to fill in. I tried lost of combos, only switching 4 and 5. I couldn’t get anything to work. I then started over with changing numbers. When ever I would change a number, lots of combos would change, and I was not having fun. Looking back, I was getting close at the end, but I didn’t know that. I spent an hour with this strategy. I put the problem aside for a few days.

I then received the hint sheet on Friday. After getting the sheet, I solved the basic solution to the POW in 20 minutes. I used the hints page that you passed out to us. The first 2 hints were my best friends. I first found what the heaviest and lightest bails weighed. Well actually I did that the other way around. I found that the lightest must be 1 and 2. (Note: 1 is always the lightest and 5 is always the heaviest, and the other numbers are in order) This is because when you combine the 2 lightest, they add up to be the lightest. Likewise the heaviest, must be bails 4 and 5. I then made this chart: I also made a chart of what I thought each bail weighed:
80 1+2
82  
83  
84  
85  
86  
87  
88  
90  
91 4+5
1 ?
2 ?
3 ?
4 ?
5 ?

Throughout the problem, I updated this chart when I found more info.

Now I wanted to find 1 and 2. I know that they add up to 80. I also know that they both can’t be 40 because you told us, and when I add 1+3 and 2+3, they would but equal the same thing. I decided to make 1 equal 39, and 2 =41. It could not be the other way around (1=41, 2=39) because then Bale 1 would not be lighter. However, it really makes no difference, but they are easier in order.

I then made these charts of possible combos (without repeating): I also "added" the possible solutions to see how "big" they would be. (ex. 1+2=3) Numbers which are big, in this simple form, would be big when I added the hay bails. I guess this would kind of be like variables. This is what I call a "weight."

1+2 2+3 3+4 4+5
1+3 2+4 3+5  
1+4 2+5    
1+5      
3 5 7 9
4 6 8  
5 7    
6      

To find these "weights," I played around. These weights also go onto the combos chart like this: They must go in order. Here is also an updated answer chart:

80 1+2(3)
82 4
83 5
84 5
85 6
86 6
87 7
88 7
90 8
91 4+5(9)
1 39
2 41
3 ?
4 ?
5 ?

Because 90 is the only combo with a "weigh" of 8, 3+5 must be that combination, because it also is the only pair with a "weight" of 8. This can also be done in reverse with 84, which has the exclusive "weight" of 4, so it must be 1+3. I can now find bale 3. It must be 82-39 (Which is Bale 1) =43. Bail 3 weighs 43 kg. Here is a recap so far:

80 1+2(3)
82 1+3(4)
83 5
84 5
85 6
86 6
87 7
88 7
90 3+5(8)
91 4+5(9)
1 39
2 41
3 43
4 ?
5 ?

I can now find 5. It must be 90(3+5)-43(#3)=47. So, we now have 5. We can also get combinations now. 83, with a "weight" of 5, must be 1+4 or 2+3. I will try 2+3 first. 41+43=88, which is not 83. This means that 83 must be 1+4. However we don’t know bale 4, but we do know that it is 83-39(combined #-1 bale known), which equals 44. Ta-da. Bale 4=44.

80 1+2(3)
82 1+3(4)
83 1+4(5)
84 5
85 6
86 6
87 7
88 7
90 3+5(8)
91 4+5(9)
1 39
2 41
3 43
4 44
5 ?

Next up is 84. This must be 2+3, because our other 5 "weight" is already solved at 83. It works when I check it. 41+43=84.

Next up, we do the same finding our 6 and 7 "weights." 85 can be 1+5 or 2+4. Lets check, 2+4, 41+44=85 Check. Next up is 86, which must be 1+5. Wait we don’t know 5, but we know that it is 86-39+47. Bada-Bing Bale 5=47. We now know all the bails. Lets just finish up, by doing 87 and 88. They are either 2+5 or 3+4. 41+47=88, so 88 is 2+4. This leaves 87 to be 3+4 or 43+44=87. There we go, POW solved.

Work

Solution

80 1+2(3)
82 1+3(4)
83 1+4(5)
84 2+3(5)
85 2+4(6)
86 3+4(6)
87 2+5(7)
88 2+5(7)
90 3+5(8)
91 4+5(9)
1 39
2 41
3 43
4 44
5 47

I then checked all of the weights to make sure they can work. (ex. 1+2 is 39+41+80 Check) I know none can repeat because all of the combs are listed here, and all work with none left over.


Extension

  1. Are there more weights?
  2. Can it be done faster?

Let’s start with 2. I think my way is fast to find answers. It just takes a lot of writing. Someone told me that you can do it with averages of all the numbers, but I know no details. It might not be faster.

1. I don’t think I can find more answers. This all works with a delicate balance of numbers, where if you change one, you change 5 of them. There also is an acceptable range where nothing else would fit. I don’t think there are any more whol numbers that could fit. I also did lots of combos by guess and check where I couldn’t find an answer.


Evaluation

Not necessary to do.