+0

# PLS HELP!!!

0
76
1

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 $C^{3}(n) = 16$.

Feb 20, 2023

#1
0

To solve this problem, we need to apply the function C three times to n, and then check if the result is equal to 16. That is, we need to find all n such that C(C(C(n))) = 16.

Let's start with an arbitrary positive integer n and apply the function C to it. If n is even, then C(n) = n/2. If n is odd, then C(n) = 3n + 1. Let's consider these two cases separately.

Case 1: n is even

If n is even, then C(n) = n/2. Applying the function C again, we get:

C(C(n)) = C(n/2) = (n/2)/2 = n/4

Applying the function C a third time, we get:

C(C(C(n))) = C(n/4) = (n/4)/2 = n/8

Therefore, in this case, we need to solve the equation n/8 = 16, which gives n = 128.

Case 2: n is odd

If n is odd, then C(n) = 3n + 1. Applying the function C again, we get:

C(C(n)) = C(3n + 1) = (3n + 1)/2, since 3n + 1 is even

Applying the function C a third time, we get:

C(C(C(n))) = C((3n + 1)/2) = (3n + 1)/4

Therefore, in this case, we need to solve the equation (3n + 1)/4 = 16, which gives n = 63.

So the two solutions to the equation C(C(C(n))) = 16 are n = 128 and n = 63.

Feb 20, 2023