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 5, 2023

#1**0 **

Here are the possible values of n such that f(f(f(n))) = 16:

n = 7 n = 13 n = 25 n = 65

To solve this problem, we can use the following steps:

First, we need to find all the values of n such that f(n) = 16. This can be done by checking all the possible values of n, and using the definition of f(n).

Once we know all the values of n such that f(n) = 16, we need to find all the values of n such that f(f(n)) = 16. This can be done by using the definition of f(n) recursively.

Finally, we need to find all the values of n such that f(f(f(n))) = 16. This can be done by using the definition of f(n) recursively again.

The following is a detailed explanation of each step:

Step 1:

We can find all the values of n such that f(n) = 16 by checking all the possible values of n, and using the definition of f(n).

The definition of f(n) is as follows:

f(n) = n/2 if n is even, f(n) = 3n + 1 if n is odd.

If n is even, then f(n) = n/2. Therefore, the possible values of n such that f(n) = 16 are 16, 32, 64, ...

If n is odd, then f(n) = 3n + 1. Therefore, the possible values of n such that f(n) = 16 are 15, 41, 67, ...

Step 2:

Once we know all the values of n such that f(n) = 16, we need to find all the values of n such that f(f(n)) = 16. This can be done by using the definition of f(n) recursively.

The definition of f(n) recursively is as follows:

f(n) = n/2 if n is even, 3f(n/2) + 1 if n is odd.

If n is even, then f(f(n)) = f(n/2). Therefore, the possible values of n such that f(f(n)) = 16 are 16, 8, 4, ...

If n is odd, then f(f(n)) = 3f(n/2) + 1. Therefore, the possible values of n such that f(f(n)) = 16 are 15, 41, 67, ...

Step 3:

Finally, we need to find all the values of n such that f(f(f(n))) = 16. This can be done by using the definition of f(n) recursively again.

The definition of f(n) recursively is as follows:

f(n) = n/2 if n is even, 3f(n/2) + 1 if n is odd.

If n is even, then f(f(f(n))) = f(f(n/2)). Therefore, the possible values of n such that f(f(f(n))) = 16 are 16, 8, 4, ...

If n is odd, then f(f(f(n))) = 3f(f(n/2)) + 1. Therefore, the possible values of n such that f(f(f(n))) = 16 are 15, 41, 67, ...

The possible values of n such that f(f(f(n))) = 16 are 7, 13, 25, and 65.

Guest Jun 5, 2023