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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

Would appreciate any help with this one.

It's from a swedish book, so I'll translate, even though we all speak swedish now don't we.

Translation:

Balancing act.

The arms on these scales are divided into sections. A weight that is placed two sections away from the middle will become twice as heavy as a weight that is placed one section away. Will you be able to place the weights so that the scale will become balanced?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi fillyfog;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi fillyfog,

The numbers written on the block, is it the weight like 2kg, 3kg etc., or are they all equal weights?

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

Don't think it matters what kind of weights they are, the numbers just have to be the same on each side so that scale is balanced.

Can only put each of the number onto the scale once.

If you put down a 2 that is one section away from the middle, it will remain a 2.

If you put down a 2 that is two sections away from the middle, it will become a 4.

Btw, I have a picture of the solution from the back of the book. I will post it in a bit, just thought I would give some of you a chance to solve it without it.

But the thing is, even with the solution, I haven't been able to work out how they came to the solution.

*Last edited by fillyfog (2011-08-07 17:50:08)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi;

Don't you have to use all the 8 weights at once?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

All weights have to be used.

There are 8 slots, and 8 numbers.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi;

Then I will have to stand with post #2.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

This is the solution from the book.

Could any of you explain how they came to this solution?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

If that is a solution then I am computing the weights on the second level differently.

But it is easy to see what he did. His diagram is somewhat misleading.

27 + 10 + 16 = 7 + 6 + 4 + 18 + 18

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

Could you explain how you got each of the numbers?

*Last edited by fillyfog (2011-08-07 18:26:35)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi;

You mean a mathematical method to solve the problem. That is not likely.

You have to solve a linear diophantine equation in 8 unknowns and in integers. There is no hand method to do that.

In addition there are now 184 solutions!

I can show you what the answer means.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi,

I get 2 solutions:

9 5 8 4 7 2 6 3

9 7 8 4 2 5 6 3

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

Just if you could explain how you got your numbers: 27 + 10 + 16 = 7 + 6 + 4 + 18 + 18

For example I understand that you got the 10 from 5 + 5, because the 5 was placed two sections away from the middle.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

That is easy:

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi,

Any of you saw #12?

bobbym,

How is 3 terms on LHS and 5 on RHS?

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi gAr;

Yes, I did. I am waiting to handle the questions that the op has before going into deeper territory. The answer I believe is 184 solutions. We will prove or disprove that as soon as he is satisfied.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Okay,

I'll take a break. See you all.

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi gAr;

Look at the diagram of his solution. There are only 3 weights on the left side of the pivot. There are 5 weights on the right side. Just go straight up from the weights to the top line to get the multiples. That is why his book answer works!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

Isnt' the 4 weight considered to be on the left side of the scale.

Meaning the scale which the 8 and 4 are on, is considered to be on the left side of the main scale.

So shouldn't the 1*4 be on the left side instead of the right.

I'm not a math wizard. But I am starting to understand how they came to the solution.

The 9*3 you get from the 9 being three sections away from the middle and so on.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Then you can see if you put that four on the left, there answer is not solution. Try it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

Thanks a ton for explaning the math in the puzzle for me.

I guess the scale is just a bit weird.

But I understand how they got to the solution in the book now.

Also, now I know where to post all my math puzzles.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi;

You are welcome to do so! I too do not think their drawing makes any sense.

Hi gAr;

I will take a bit of a break to eat.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**fillyfog****Member**- Registered: 2011-08-07
- Posts: 9

Going back and checking the book's solution with what I now know, I think gAr's solution 9 7 8 4 2 5 6 3 is the one they should have gone with in the book instead of the 9 5 8 4 7 2 6 3.

If you go with how physics work, there is actually 4 weights on each side of the main pivot. With that in mind gAr's solution 9 7 8 4 2 5 6 3 works, which adds up to 57 on each side.

Either way, thanks again for all of your help.

There's a ton of math puzzles in the book I am solving at the moment, so I might be back to pick all of your magnificent brains.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi;

Remember post #9?

bobbym wrote:

If that is a solution then I am computing the weights on the second level differently.

When you showed the book solution, I just adjusted my method to how they were figuring the second level. I did not have their answer either.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

When the balance is in equilibrium, the sum of torque = 0

Applying that to each of the balances, I get my solutions.

These are my equations:

Name the weights from left to right a,b ...h, then

2*d = c

g = 2*h

(c+d)+2*b+3*a = e + 3*f + 4*(g+h)

Using first 2 equations, 3rd one is

3*d + 2*b + 3*a = e + 2*f + 12*h

Brute forcing for values 2,3 ... 9, we get the two solutions.

Hi fillyfog,

You're welcome!

*Last edited by gAr (2011-08-07 21:23:34)*

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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