Weighing the odd one out in a group of 12 objects using only 3 uses of a balancebeam scale The basic premise is to find the odd one out (weight-wise) in a group of 12 objects. One object has a different weight than the others. Using the balancebeam scale three times, find out which one is which. 
How would I even begin to go about this question? It's stumped me for over an hour. You are capable of weighing multiple.
 A: Label your rabbits from $1$ to $12$.
First, scale $1-4$ against $5-8$. 


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*If they are the same, the odd one is within $9-12$. In that case scale $9-11$ against $1-3$. 


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*If they are the same, $12$ is the odd one and one more try will tell
you if it's over or underweight.

*If they are different, the odd one is within $9-11$ and you know if it's under/overweight since $1-3$ are good. Just scale $9$ against $10$ now.


*If $1-4$ and $5-8$ are different (say $5-8$ heavier), you know that whenever you find out who is the odd one you will automatically know whether it's over or underweight.
Now is the tricky one
Weigh $1-3$ and $5$ together against $4$ and $9-11$ (the last three are normals).


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*If $1-3$ and $5$ go down, it can only be because of $5$ being heavier or $4$ being lighter; the last try will easily tell which (eg $5$ against $12$).

*If they go up, the odd one is amongst $1-3$ and you find out which by scaling $1$ against $2$.

*If it stays still, the odd one is amongst $6-8$ and you scale $6$ against $7$. 
