Function C is defined on positive integers as follows:
\(C(n) = \begin{cases} \dfrac n 2 & \text{if $n$ is even}, \\ 3n+1 & \text{if $n$ is odd}. \end{cases}\)
Find all n such that C^3(n)=16
Since C(n) is always odd, C(C(n)) must be a multiple of 3. Therefore, n must be a multiple of 3. We can check that C(16)=16 and C(C(16))=16, so n=16 is a solution. We can also check that C(C(C(48)))=16, so n=48 is also a solution. These are the only solutions, since C(C(C(n))) is always a multiple of 9, and there are no positive integers that are a multiple of 9 and a multiple of 3 but not a multiple of 27.
So the only solutions are 16 and 48.
