# A load of balls!



## robert@fm (Apr 28, 2016)

You have six balls, two each red, green and blue. The two balls of each colour are visually indistinguishable, but one weighs 200g while the other weighs 210g. You have a sensitive lever balance, the pans of which can each hold one or two of the balls, but you are only allowed two weighings (with three, the problem would be trivial). How can you determine, for all six balls, which is which?


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## Redkite (Apr 28, 2016)

When you say only two weighings allowed, does this count incremental removal of a ball from one of the pans?


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## Diabeticliberty (Apr 28, 2016)

robert@fm said:


> You have six balls, two each red, green and blue. The two balls of each colour are visually indistinguishable, but one weighs 200g while the other weighs 210g. You have a sensitive lever balance, the pans of which can each hold one or two of the balls, but you are only allowed two weighings (with three, the problem would be trivial). How can you determine, for all six balls, which is which?




Come on then. Tell us??????


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## robert@fm (Apr 28, 2016)

The first weighing is of one or two balls in each pan (there's obviously no point in placing one ball in one pan and two in the other, as the pan with two balls will always go down, so this doesn't tell you anything). The second weighing is a different combination of balls, which may include some from the first weighing. It must also be noted that this is a simple lever balance; a heavier weight won't make a pan go down further, either it goes down or it doesn't.


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## robert@fm (Apr 28, 2016)

OK, another puzzle in the same vein but probably easier (?):

You have 12 balls, identical in appearance but one is ether heavier or lighter than the rest, while the others are all the same weight. You also have a (simple) lever balance, the pans of which can hold as many balls as needed. In three weighings, completely identify the odd ball. (Some notation to help: Letter the balls A through to L. WX-YZ means that W and X are in the left pan, Y and Z in the right pan.)


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## Annette (Apr 28, 2016)

I have an answer. I'm just trying to work out how to put it in words...


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## Annette (Apr 28, 2016)

Red and Blue in one side, Red and green in other. 
If the weights are equal, you know you have R(h) and B or G(l) with R(l) and B or G(h). Weigh the red balls to find out which is which, hence you know which B or G you have, and hence what they are.
If the weights are not equal, the heavy side must have the R(h), so you know the Reds. (If it has the R(l), the heavy side could only have (l) + (h or l) and the other (h) + (h or l) so wouldnt be heavier). Weigh R+R against the original B+G. If B+G heavier, both of them are heavy (and their partners light). If B+G lighter, both are light (and partners heavy.) If same, then one is heavy and one light. The one that was originally with the heavy Red must be heavy and the one with the light Red must be light (otherwise the original sides not being equal wouldnt be correct.)
Umm. I'm not sure how well thats explained, but I'm fairly sure its right?


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## Diabeticliberty (Apr 28, 2016)

Annette Anderson said:


> Red and Blue in one side, Red and green in other.
> If the weights are equal, you know you have R(h) and B or G(l) with R(l) and B or G(h). Weigh the red balls to find out which is which, hence you know which B or G you have, and hence what they are.
> If the weights are not equal, the heavy side must have the R(h), so you know the Reds. (If it has the R(l), the heavy side could only have (l) + (h or l) and the other (h) + (h or l) so wouldnt be heavier). Weigh R+R against the original B+G. If B+G heavier, both of them are heavy (and their partners light). If B+G lighter, both are light (and partners heavy.) If same, then one is heavy and one light. The one that was originally with the heavy Red must be heavy and the one with the light Red must be light (otherwise the original sides not being equal wouldnt be correct.)
> Umm. I'm not sure how well thats explained, but I'm fairly sure its right?



WHAT SHE SAID√√√


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## Annette (Apr 28, 2016)

Ok, as no-one's answered the other one (I gave people some time), here goes:
Put 6 balls in one side and 6 in the other: ABCDEF-GHIJKL
one side will be heavier (with the heavy ball in). Take those balls and put 3 in one side, 3 in the other: ABC-DEF
one side will be heavier (with the heavy ball in). You have ABC, and one is the heavier.
Weigh A-B. If A is heavier, there's your answer. If B, ditto. If A and B are the same, C is the heavier.


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## robert@fm (Apr 28, 2016)

Sorry, but that doesn't work; the odd ball in this case could be heavier or lighter. 

However, you're on the right track...


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## Robin (Apr 28, 2016)

Annette Anderson said:


> Ok, as no-one's answered the other one (I gave people some time), here goes:
> Put 6 balls in one side and 6 in the other: ABCDEF-GHIJKL
> one side will be heavier (with the heavy ball in). Take those balls and put 3 in one side, 3 in the other: ABC-DEF
> one side will be heavier (with the heavy ball in). You have ABC, and one is the heavier.
> Weigh A-B. If A is heavier, there's your answer. If B, ditto. If A and B are the same, C is the heavier.


So far so good, I'd got that far, but Robert's question said one ball was either heavier or lighter than the rest, and that's what foxes me.


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## Robin (Apr 28, 2016)

Oops, Roberts reply sneaked in just before mine!


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## Annette (Apr 28, 2016)

Ah, ok, I read it that you'd know if it was heavier or lighter.


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## Annette (Apr 28, 2016)

ok, try this:

Weigh ABCD-EFGH
If they are equal then different ball is one of IJKL.
Weigh EFG-IJK. If same then L is the odd ball. Weigh against A to see if heavy or light.
If IJK is heavier, it has a heavier ball in it. Weigh I-J, if they balance, K is a heavier ball. If I or J is heavier, there’s your answer.
If IJK is lighter, it has a lighter ball, continue as above.

If EFGH is heavier than ABCD either EFGH has a heavier ball or ABCD has a lighter ball. Weigh ABE-CFL.
If same, either D is light or G or H are heavy. Weigh G-H. If same, D is light. Or, whichever is heavy is (duh) heavy.
If CFL is heavy, either F is heavy or A or B is light. Weigh A-B (continue as above).
If CFL is light, C is light or E is heavy. Weigh A-C, if same, E is heavy, else C is light.

If ABCD is heavier than EFGH, either ABCD has a heavy ball or EFGH has a light ball. Weigh AEF-BGL.
If same, either H is light or C or D is heavy. Weigh C-D to tell which.
If BGL is heavy, either B is heavy or E or F is light. Weigh E-F to tell which.
If BGL is light, either G is light or A is heavy. Weigh G-B, if same, A is heavy, else G is light.

(With some help and hindrance from the mathematician in the next door office to mine  )


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## Ralph-YK (Apr 28, 2016)

Sorry.  Surely Annette's first answer to the 12 ball question works whether one ball is heavier or lighter.  If not, I've not understood one work of it.


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## Annette (Apr 28, 2016)

Ralph-YK said:


> Sorry.  Surely Annette's first answer to the 12 ball question works whether one ball is heavier or lighter.  If not, I've not understood one work of it.


Yes, but only if you know whether it is heavier or it is lighter. If you know, then it works (group 1 is heavier, therefore it contains the different ball.) If you only know its different, it doesnt (group 1 is heavier, therefore it could contain a heavier ball. Or Group 2 could contain a lighter ball.)
Any clearer?


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## Ralph-YK (Apr 28, 2016)

Yes.  *mutters. goes and sits in the corner again*


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