# New Brain Teaser



## BoaterDan (Jul 1, 2005)

Ok folks.

Imagine you have 8 small balls. They are all identical in every way except that one of them weighs slightly more than the other 7. The weight difference is too small for you to detect without an instrument. The only thing you have besides the balls is a balance scale - the type where you put items on both sides and it tips towards the heavier side. What is the least amount of weighings you have to make to find the odd ball, and what is your procedure?

All the information you need is contained in the above paragraph.

Good luck.


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## CamperAndy (Aug 26, 2004)

Weigh 2 at a time.

The least number would be 1 the most would be 4.


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## larry (Oct 19, 2004)

Number of weighs is TWO

First weigh three and three

If they are equal, then next weigh one and one ( one must be heavier )

If they are not equal, weigh only two on the heavy side.

If one is heavier , job complete.

If they are equal, then the ball that you did not weigh is heavier


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## PDX_Doug (Nov 16, 2004)

My first thought is: 'Oh, goodie, another brain teaser!









Then Larry comes along and gets the answer before I even get a chance to weigh in!









Larry's idea looks spot on, and I can't come up with a way to do it in a single step. Not that I will not keep trying.

Larry, you are 'da man!









Happy Trails,
Doug


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## CamperAndy (Aug 26, 2004)

PDX_Doug said:


> My first thought is: 'Oh, goodie, another brain teaser!
> 
> 
> 
> ...


I must be chopped liver but I don't see how Larry's procedure is the most expeditious.

If you pick any 2 balls at random and weigh them one on each side of the scale, you will find the heavy one in from one to four weighings.

Using Larry's procedure will also result in a minimum of 2 weighings and as many as 4.


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## gregjoyal (Jan 25, 2005)

CamperAndy said:


> I must be chopped liver but I don't see how Larry's procedure is the most expeditious.
> 
> If you pick any 2 balls at random and weigh them one on each side of the scale, you will find the heavy one in from one to four weighings.
> 
> ...


I like your method, but you should mention that you pick two from the eight, then two from the remaining six, etc...


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## PDX_Doug (Nov 16, 2004)

CamperAndy said:


> Using Larry's procedure will also result in a minimum of 2 weighings and as many as 4.


Andy,

Maybe I am missing something, but the way I read Larry's procedure, two weighings will always yield the answer, even if only by inference.

With your procedure, you have a 25% chance of doing it in one weighing, and a 50-50 chance of finding the heavy ball in two weighings.









Happy Trails,
Doug


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## CamperAndy (Aug 26, 2004)

PDX_Doug said:


> CamperAndy said:
> 
> 
> > Using Larry's procedure will also result in a minimum of 2 weighings and as many as 4.
> ...


I worked this some and will agree that you can always derive the answer in 2 weighings. I did not understand Larry's procedure at first so I will try to illustrate it more.

There are 8 balls numbered 1 to 8

Weigh 1,2 and 3 on one side and 4, 5 and 6 on the other.

If they are equal then weigh balls 7 and 8 for the heavy one.

If they are not equal take the 3 balls from the heavy side, say 4, 5 and 6 and weigh just 2 of them, say 4 and 5. You will find the heavy one on the scale or it will be the one not weighed.


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## PDX_Doug (Nov 16, 2004)

Exactly!









Whew! I really wasn't looking forward to another protracted argument!*
But then, we have not heard from Thor yet! (Just kidding!)









Happy Trails,
Doug

* Actually, I was... but you didn't hear it from me!


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## BoaterDan (Jul 1, 2005)

Yeah, Larry's got it.

I guess the question should read "what's the least number of weighings you have to make to be guaranteed of finding the odd ball".


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## Thor (Apr 7, 2004)

I just read this one, sorry I am late. Larry has nailed it.









Thor


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## HootBob (Apr 26, 2004)

Ok my brain is fried









don


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## wolfwood (Sep 19, 2005)

I am sooo disappointed. Geez - only 1 page - no one even called in the Science Guy! So you know which is heaviest - but doesn't anyone care where that ball was made? Is it a Made in the USA ball? Was it outsourced to China?

Surely there's more fun to be had with this - THOR, WHERE ARE YOU ??????


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## nynethead (Sep 23, 2005)

No, it was manufactured in china, but outsourced to india and paid for by an american


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## PDX_Doug (Nov 16, 2004)

wolfwood said:


> Surely there's more fun to be had with this - THOR, WHERE ARE YOU ??????


Thor here, wolfwood.

In case you have not figured it out, this whole PDX_Doug thing is just a ruse perpetuated by my alter ego!









Happy Trails,
Doug.. er... Thor

(Yeah, that's the ticket!)


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## Morrowmd (Feb 22, 2005)

OK, I will perpetuate the thread.

I was thinking you would weigh 4 & 4, take the balls on the heavy side and weigh 2 & 2, again take the heavy side and weigh 1 & 1.

This would guarantee the right answer in 3 steps.

Am I right or am I right?

OK, upon further review I see how Larry's procedure could get the answer in 2 steps, or 3 at the most. So technically speaking that is the best answer because you could get it in 2 steps.

I don't think there is a way to get it in 2 steps guaranteed, though.

-Matt


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