Difficult Brainteaser type question (1 Viewer)

[Seth--L]

I'll be impressed if anyone can get this. I sure can't:

There are 12 identical looking stones. Inside one of the stones is a diamond that you want to find. 11 of the stones weigh exactly the same, and the one with the diamond weighs either slightly more or slightly less than the others. You are given three balance scales (like the kind that Lady Justice holds), and you may only use each scale once. When using the scales, you may NOT put one stone on at a time, or put a bunch on and then remove stones one at a time. So, how do you find the stone with the diamond, and how do you determine that it is lighter or heavier than the other 11.

 

[James T]

Are you sure you cannot put one stone at a time? And are you sure it doesn't specify if it's heavier or lighter? I heard it before, but it specified it was heavier.

My solution was break it into 6 and 6. The heavier one is put aside. Break it into 3 and 3 and the heavier is put aside. Pick any two and balance. If it balances perfect, the remaining one is the one with the diamond. If one tips as being heavier, then that's the diamond.

 

[Tom Vodhanel]

1) put 3 on each side of scale one.

if they measure the same, you know it is in one of the other 6.

2)put the other 6 on scale two.

3)whichever side measures heavier...take any 2 of the 3 stones and weigh them on scale three(edit---just noticed you can't weigh one at a time.

If they are the same, you know it was the only stone from the heavier side of scale 2.

TV

 

[Seth--L]

Yes, it's up to you to figure out if it's lighter or heavier.

 

[StephenK]

Tom and James,

Your solutions only work if you already know whether the diamond rock is heavier or lighter.

Seth,

Are you sure that's not part of the starting info? Cause now, I'm stumped

 

[James T]

I did some research and his wording is right, but no one has an answer when it's missing the weight.

Another way was 4 vs 4. If it balances, then use the remaing four and 2 v 2. Then you're stumped b/c you don't know if it's heavy or lighter.

And I think I misread the "ONE stone at a time" as you are not allowed to put 1 v 1...which would make this riddle impossible.

 

[Seth--L]

Yes.

 

[James T]

My afternoon was shot b/c of this riddle. Damn this thread.

Personally, since I haven't seen an answer, I think this is the new version of the GRY riddle. For those that don't know, the GRY riddle is impossible to answer if told incorrectly, but has been told incorrectly for decades.

 

[Chris_Morris]

Well, I'm not going to spend the hours it takes to solve this riddle. So I'll just post a link to the answer (the site also has alot of other brainteasers):

Gray Watson's Brain Teasers



Chris

 

[BrianW]

The key is to remember that the scales don't just yield a binary result of equal or not equal. They yield one of three results, equal, heavy-left, or heavy-right. By using this trinary system, you can eliminate more than half the stones with a single weighing. By strategically arranging the stones so that each of the three outcomes points to a different set of stones, you'll narrow it down much more quickly.

For instance, if you weigh three stones on the left side and three stones on the right side, and it is heavy left, then in your next weighing, you can eliminate not just half, but two-thirds of the stones.

How?

For the next weighing, do the following:

* Take a stone from each side and set it aside. These two will be the cast-off stones.

* Take a stone from the left side and put it in the right side, and take a stone from the right side and put it in the left side. These two will be the exchange stones.

* Leave the other two stones on each side where they are. These two will be the untouched stones.

Now weigh them again. If the left side is still heavy, then you know that the different stone – whether heavy or light – remains unmoved in the scale, so it must be one of the two untouched stones.

If the right side is heavy, then you know that the different stone has switched sides, so it must be one of the two exchange stones.

If the two sides are equal, then you know that the different stone has moved off the scale, so it must be one of the two cast-off stones.

So in one additional weighing, you've eliminated two-thirds of the stones. From here, one more weighing will reveal the different stone.

But when starting with twelve stones, should you ever weigh three against three (or six against six)? Probably not, since that would imply a binary – rather than trinary – system. But it's a good example to get your mind going. Don't forget that reintroducing stones known not to contain the diamond can reveal whether the diamond stone is heavier or lighter than the rest of the stones. Is this something you should do? Or would that just be a wasted turn at the scales? Well, the answer is: it depends!