$\textbf{Heavier Ball}$
There are $12$ balls of which $11$ weigh the same and the other one is heavier. What is the minimum number of weighs required to find this heavier ball using a balance?
$\textbf{Solution}$
Three weighs are sufficient. Dividing the balls equally into three groups of $4$ each. Weigh the first group against the second group. If their total weight is different, then the abnormal ball is in the heavier group. Otherwise, if they weigh the same, the abnormal ball is in the third group. Nevertheless, we now have a group of $4$ balls to check. Next, pair them one to one, and it takes at most another two weighs to find the heavier ball.
$\textbf{Note}$
Problems like this appear frequently in different tests. The number of balls varies, but if we know the abnormal object is heavier or lighter, then $n$ weighs is sufficient to find this outlier among up to $3^n$ objects.
If we only know the outlier weighs differently (can be either heavier or lighter), then the problem is a harder. See # 4664.