Function $C$ is defined on positive integers as follows:Find all $n$ such that $C^{3}(n) = 16$.
\(\[C(n) = \begin{cases} \dfrac n 2 & \text{if $n$ is even}, \\ 3n+1 & \text{if $n$ is odd}. \end{cases}\]\)
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.
