Function $C$ is defined on positive integers as follows:

Find all $n$ such that $C^{3}(n) = 16

\\(\[C(n) = \begin{cases} \dfrac n 2 & \text{if $n$ is even}, \\ 3n+1 & \text{if $n$ is odd}. \end{cases}\]\)

Guest Jun 6, 2023

#1**0 **

We have that C(C(C(n)))=16. If n is even, then C(n) is odd, so C(C(n)) is even, so C(C(C(n))) is a multiple of 9. However, 16 is not a multiple of 9, so n cannot be even.

If n is odd, then C(n) is even, so C(C(n)) is odd, so C(C(C(n))) is a multiple of 3. Since 16 is a multiple of 3, we have that C(C(n))=16. Then C(n)=3n+1=16, so n=5.

Therefore, the only solution is n=5.

Guest Jun 6, 2023