Weighing the odd one out in a group of 12 objects using only 3 uses of a balancebeam scale

Question

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.

Answer 1

Label your rabbits from $1$ to $12$.

First, scale $1-4$ against $5-8$.

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

  • 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).

  • 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 $7$.