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# Weighing V.

23 replies to this topic

### #21 WizardOfSpeedAndTime

WizardOfSpeedAndTime

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Posted 06 June 2008 - 08:04 PM

Maybe this is too simple...maybe I'm missing something...you don't need to measure 4 balls at once on 2 scales...

I'd keep it more simple since all light balls weigh the same and all heavy balls weigh the same.

1) Weigh one white ball and one white ball - one on each scale.
- we find out the weight of the light ball
- we find out the weight of the heavy ball
2) Weigh one red ball and one blue ball one on each scale.
- we find out the red ball is heavy or light.
- the other red ball is the opposite
- we find out the blue ball is heavy or light.
- the other blue ball is the opposite

This does not consider the case where your second weighing has equilibrium. Then you don't know if they're heavy or light, just that they're the same.
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### #22 mayiko

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Posted 04 July 2008 - 05:44 PM

I got almost the same thing as the spoiler except for the 2nd weighing, I'd weigh the red and blue balls from the first weighing against the two white balls because from the first weighing, we can deduce that those red and blue balls are both light, both heavy, or the one on the heavier side is heavy and the one on the lighter side is light. You can find out which one applies by weighing it against the two weight balls, and then you know everything.
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### #23 loslloydos

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Posted 12 September 2008 - 12:01 PM

Spoiler for Another option

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### #24 kyky

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Posted 02 October 2009 - 12:00 PM

easy.

first balls: a b
second balls: c d
third balls: e f

now we weight, b and c.. d and e only..
if b = c, d = e then we can conclude b c f / a d e
if b =/c , d =/e then we can conclude b d f / a c e ( if b and d heavier )
if b = c, d =/e then b c e / a d f.

you dont have to know which one heavier or lighter, you can simply group them up with simple measurement
question?

Edited by kyky, 02 October 2009 - 12:02 PM.

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