A function `f:ZrarrZ` is defined as `f(n)={{:(n+1,n in " odd integer"),((n)/(2),n in "even integer"):}` . If k `in` odd integer and `f(f(f(k)))=33`, then the sum of the digits of k is

Options

A. 7
B. 5
C. 9
D. 9

Answer

To solve the problem, we need to analyze the function \( f \) defined as follows:

\[ f(n) = \begin{cases} n + 1 & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases} \]

We are given that \( k \) is an odd integer and \( f(f(f(k))) = 33 \). We need to find the sum of the digits of \( k \).

### Step 1: Calculate \( f(k) \)

Since \( k \) is odd, we use the first case of the function:

\[ f(k) = k + 1 \]

### Step 2: Calculate \( f(f(k)) \)

Now \( f(k) = k + 1 \), which is even (since \( k \) is odd). Thus, we use the second case of the function:

\[ f(f(k)) = f(k + 1) = \frac{k + 1}{2} \]

### Step 3: Calculate \( f(f(f(k))) \)

Now we need to find \( f(f(f(k))) \). Since \( f(f(k)) = \frac{k + 1}{2} \), we need to determine if this result is odd or even.

1. If \( k + 1 \) is even, \( \frac{k + 1}{2} \) is an integer.
2. We need to check if \( \frac{k + 1}{2} \) is odd or even.

Since \( k \) is odd, \( k + 1 \) is even, so \( \frac{k + 1}{2} \) is an integer and we can analyze it further.

Now, we need to check the parity of \( \frac{k + 1}{2} \):

- If \( k + 1 = 2m \) for some integer \( m \), then \( \frac{k + 1}{2} = m \), which can be either odd or even depending on \( m \).

### Step 4: Determine \( f(f(f(k))) \)

Assuming \( \frac{k + 1}{2} \) is even (which it is, since \( k + 1 \) is even), we apply the second case of the function:

\[ f(f(f(k))) = f\left(\frac{k + 1}{2}\right) = \frac{\frac{k + 1}{2}}{2} = \frac{k + 1}{4} \]

### Step 5: Set up the equation

We know that:

\[ f(f(f(k))) = 33 \]

Thus, we have:

\[ \frac{k + 1}{4} = 33 \]

### Step 6: Solve for \( k \)

Multiplying both sides by 4 gives:

\[ k + 1 = 132 \]

Subtracting 1 from both sides gives:

\[ k = 131 \]

### Step 7: Find the sum of the digits of \( k \)

Now we need to find the sum of the digits of \( k = 131 \):

\[ 1 + 3 + 1 = 5 \]

### Final Answer

The sum of the digits of \( k \) is \( \boxed{5} \).