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[Magma]

[Ivy Set] #164 Weighs and Means

[fretty]

On your small balance, the second level of weights is set differently to the card so you might have to use the left hand number as your answer for weight number 3. Don't know whether you're doing this already.

I can't spot any errors with copying the large diagram but haven't checked the formulas yet.

[Magma]

I didn't put the ticks in the small weight diagram, but the spacing is right - hence the 2kg one tick away balancing the 1kg two ticks away.

[fretty]

No I was just saying that when you copy your answer from your spreadsheet you may cross reference it with the card. So with weight 3 being right of the join, you could have copied the answer that is right of the join on the sheet when your answer is left of the join.

[Magma]

Okay, I've updated the XLS file with a new version - the weights you need are numbered #1, #2, etc. and each page has a table that says what the weights the puzzle asks for are currently set to, to avoid confusion. Is that what you meant?

[CRHB]

Given that it has to be the lowest set of weights, I tried to minimise the most distant and got something different to above. According to the site it is still did not right though Any ideas?

[Magma]

I think the key to solving this is working out the minimum weights that balance the top bar. We have this setup:

......5...........2.......4
. _ _ _ _ _ | _ _ _ _ _ _
|.......................|...........|
|.......................|...........|
A......................B..........C

Remember the force acting on the pivot is equal to the mass x distance, so on the left:

5A acts downwards.

On the right:

2B and 6C acts downwards.

So we have the equation 5A = 2B + 6C, and we need the minimum.

I thought along these lines:
The bottom right set of weights mninmum weight is 1 on the four to the right, balanced by 10 on the left (10 vs 1x1 + 2x1 + 3x1 +4x1)

This makes the next set of weights up to be 14 on the left, which minimum balance is 6 on the far right and 2 on the near right (As each weight has to be at least 1, and integer values)

This means the weight acting at position B as labelled above has to be a minimum of 22kg. So the minimum configuration of the right-hand-side is (as far as B is concerned):

44 + 6C

Similarly, working up the tree of weights to position C gives a minimum value for C as 9kg. This would make the minimum right-hand-side force 44 + 54 = 98

The minimum configuration for A is 13, which makes the minimum left-hand-side configuration

65

This leaves the left side too light by force 33. Since this is not divisible by 5, some jiggery-pokery is needed on the right side too. Anyone know offhand which integer solution for 5A = 2B + 6C has the lowest value for A + B + C given that A >=13, B >= 22 and C >= 9?

[Austin]

The right hand side's force must be divisible by 5. I can get this as low as 100 but my solution was rejected. There must be a lighter way to do the right hand side, keeping the same downward force (as the left side's weights will have to add up to 20 whatever else happens, otherwise our attempts would have worked).

So far i have 20 Kg on the left and 32 Kg on the right.

[Langley Moor]

I get the same Austin, 20kg on the right hand side producing a moment of 100, and a total moment of 100 from the right hand side produced by 23kg at the secondincrement (moment of 46) and 9kg at the sixth increment (moment of 54). However, not accepted by the website. Anyone fancy writing to mind candy, as I can't be bothered this evening? If not I'll do it tomorrow

[Stratman]

Has anyone confirmed whether you have to use 25 unique weights (ie 1kg,2kg,3kg,4kg......25kg) as in Erich Friedman's usual weight puzzles? I have tried a couple of solves with minimum weights (ie lots of 1s and 2s etc) with no luck.

link
http://www.stetson.edu/~efriedma/weight/

[Austin]

[Phill4269]

I have tried several solutions without success.

One question is what do they mean by

[Stratman]

Thanks Austin - good point.
I also had a solve with 100 each side...
Left 20x5
Right 23x2 and 9x6
...with no luck
I cant see how it can be done with less.
Stumped!

[gagravarr]

[TopGun2]

I'm struggling to see how you can get the right hand side to make 100. I can see the 9*6=54 but I can't get 23*2 I'm stuck with 22 or 25 any hints?


Secondly, it might be worth trying to set this out in a set of simultaneous equations & solving by matrices. I'll have to dig out my study notes to revise how!