Wife came home and told me this one. It’s not easy (though it looks it at first.)
You have 12 bricks and one of those justice style scales:
They all look the same, except one brick’s weight is different (maybe heavy, maybe light.) Figure out the odd brick in 3 weighing attempts.
I got it.
Answer below so as not to spoil it.
Four against four. If one is heavier, then two against two and then one against one. If they are the same, then weigh the other four as a two against two and… damn, it could be heavier or lighter. Nevermind.
You sure? I said that like 5 times. 
Edit: You have to figure out multiple paths if you know what I mean. You’ll solve it one way, and then you’ll notice that it won’t work another.
[quote=“Jaboney”]I got it.
Answer below so as not to spoil it.[/quote]
It might be light OR heavy, that is unknown.
Ok. Assuming it’s light.
Don’t matter, I broke that bad boy.
Where’s my Crackerjack prize?
[quote=“Jaboney”]Ok. Assuming it’s light.
Don’t matter, I broke that bad boy.
Where’s my Crackerjack prize?[/quote]
You can’t assume it’s light. That makes it a lot easier. In your solution.
[quote]Weigh all 12, six on each side. The side that rises has the light brick.
Weigh those six, three to a side. The side that rises has the light brick.
Pick two of those three, weigh them one to a side. If one rises, it’s light. If balanced, it’s the third brick.[/quote]
This wouldn’t work because when you get to your last step, if it’s not balanced, you don’t know which one it is.
Note: Actually, the very first step is flawed.
What are you talking about??? Ok, I see. I assume we knew if it was heavy or light. hmmm…
I did the same thing. 
Weigh four on each side. (1a) If balanced, weigh two of the remaining four. (1b) If balanced, it must be one of the two remaining. Take one and put it on one side and one of the first ones weighed and put it on the other side. The original brick is standard weight, so the scales will show you if the other brick is light or heavy.
(1b) If not balanced, weigh one against an original brick. You know which side was heavier before, so you now know if the brick remaining is light or heavy (if the scales tip), or if the one removed is light or heavy (because the scales will now be balanced).
OK, I’m working on the other scenario now …
If you get it, be careful not to spoil it for the others.
The force is strong in this one.
You’re missing a key point that putting your thoughts on paper might reveal.
Oh, wait! I don’t have to find the brick and whether it’s light or heavy - just find the odd brick, right?
That’s right. I’ll give another hint. There basically ends up being 3 paths to success. Two of these paths tells you which brick and the weight. One path just tells you which bring.
Is there a proper note taking method for figuring out problems like this? Mine ended up as a bunch of scribbles with scales and pictures.
That’s right. I’ll give another hint. There basically ends up being 3 paths to success. Two of these paths tells you which brick and the weight. One path just tells you which bring.[/quote]
I take that back. You will always find out whether it is heavier or lighter.
There are only a few ways to evenly divide the bricks for the first balancing: {6,6}, {4,4,4}, {3, 3, 3, 3}, or {2, 2, 2, 2, 2, 2}. Since nobody has so far had much luck with the groups of 4s, maybe another of these groups could be a possibility.
I think I have seen this before but the best I can do tonight with certainty is 4 weighs. If you knew whether it was either light of heavy then 3 is straight forward.
You have to group in 4’s initially to minimise the chances.
Are the rules definitely that there are only 3 weighs, or is it that you can only weigh each brick 3 times?
Just 3 attempts.
It’s the small blue brick and it’s lighter than those big red ones.
Is it only a little bit heavier/lighter, or a lot? Or don’t we know? Coz if it’s a lot heavier/lighter, you could take an educated guess by feel before you weighed.
Or - post the question on a web forum asking for other peoples solutions. Then, when they have a solution, use this as your answer. That way, you wouldn’t need 3 weighing attempts.
Oh, I forgot to mention that the bricks are encased in a glass box and you have to use robotic arms to move them around thus nullifying the ability to judge weight based on feel.
Well, you would need the correct solution for each new scenario, so it’s necessary to understand how the three steps work.
Close but no toe cheese.