A function f from integers to integers is defined as `f(x)={{:(n+3",",nin"odd"),(n//2",",nin"even"):}` suppose `kin` odd and `f(f(f(k)))`=27 then the sum of sigits of k is

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the problem step by step, we will analyze the function \( f \) and how it behaves when applied multiple times.

### Step 1: Understand the function \( f \)
The function \( f \) is defined as follows:
- If \( n \) is odd, then \( f(n) = n + 3 \)
- If \( n \) is even, then \( f(n) = \frac{n}{2} \)

### Step 2: Start with \( k \) being odd
Given that \( k \) is odd, we can apply the function \( f \) to \( k \):
\[ f(k) = k + 3 \]
Since \( k \) is odd, \( k + 3 \) will be even.

### Step 3: Apply \( f \) again
Now we apply \( f \) to \( f(k) \):
\[ f(f(k)) = f(k + 3) \]
Since \( k + 3 \) is even, we use the even case of the function:
\[ f(k + 3) = \frac{k + 3}{2} \]

### Step 4: Apply \( f \) a third time
Next, we apply \( f \) to \( f(f(k)) \):
\[ f(f(f(k))) = f\left(\frac{k + 3}{2}\right) \]
Now we need to determine if \( \frac{k + 3}{2} \) is even or odd.

### Step 5: Analyze cases for \( \frac{k + 3}{2} \)
1. **Case 1:** Assume \( \frac{k + 3}{2} \) is even:
\[ f\left(\frac{k + 3}{2}\right) = \frac{\frac{k + 3}{2}}{2} = \frac{k + 3}{4} \]
We know from the problem that \( f(f(f(k))) = 27 \):
\[ \frac{k + 3}{4} = 27 \implies k + 3 = 108 \implies k = 105 \]

2. **Case 2:** Assume \( \frac{k + 3}{2} \) is odd:
\[ f\left(\frac{k + 3}{2}\right) = \frac{k + 3}{2} + 3 \]
Setting this equal to 27 gives:
\[ \frac{k + 3}{2} + 3 = 27 \implies \frac{k + 3}{2} = 24 \implies k + 3 = 48 \implies k = 45 \]

### Step 6: Verify the cases
- For \( k = 105 \):
- \( f(105) = 108 \) (even)
- \( f(108) = 54 \) (even)
- \( f(54) = 27 \) (even)

- For \( k = 45 \):
- \( f(45) = 48 \) (even)
- \( f(48) = 24 \) (even)
- \( f(24) = 12 \) (even)

Since \( k = 45 \) does not satisfy the condition \( f(f(f(k))) = 27 \), we conclude that \( k = 105 \) is the only valid solution.

### Step 7: Find the sum of the digits of \( k \)
Now we calculate the sum of the digits of \( k = 105 \):
\[ 1 + 0 + 5 = 6 \]

### Final Answer
The sum of the digits of \( k \) is \( \boxed{6} \).