A function `f` from integers to integers is defined as
`f(n)={(n+3",",n in odd),(n//2 ",",n in even):}`
Suppose `k in odd and f(f(f(k)))=27.` Then the value of k is ________

Explanation

To solve the problem, we need to find the integer \( k \) such that \( f(f(f(k))) = 27 \) where \( k \) is an odd integer. The function \( f \) is defined as follows:

- If \( n \) is odd, \( f(n) = n + 3 \)
- If \( n \) is even, \( f(n) = \frac{n}{2} \)

### Step-by-step solution:

1. **Start with the function definition**:
Since \( k \) is odd, we can apply the function \( f \):
\[ f(k) = k + 3 \]
Here, \( k + 3 \) is even (since the sum of an odd number and an even number is even).

2. **Apply the function again**:
Now, we need to find \( f(f(k)) \):
\[ f(f(k)) = f(k + 3) = \frac{k + 3}{2} \]
This is because \( k + 3 \) is even.

3. **Apply the function a third time**:
Next, we apply \( f \) again:
\[ f(f(f(k))) = f\left(\frac{k + 3}{2}\right) \]
We need to determine whether \( \frac{k + 3}{2} \) is odd or even. Since \( k \) is odd, \( k + 3 \) is even, and thus \( \frac{k + 3}{2} \) is an integer.

- If \( \frac{k + 3}{2} \) is even, then:
\[ f(f(f(k))) = \frac{\frac{k + 3}{2}}{2} = \frac{k + 3}{4} \]
- If \( \frac{k + 3}{2} \) is odd, then:
\[ f(f(f(k))) = \frac{k + 3}{2} + 3 = \frac{k + 3 + 6}{2} = \frac{k + 9}{2} \]

4. **Set up the equation**:
We know that \( f(f(f(k))) = 27 \). We will consider both cases.

**Case 1**: \( \frac{k + 3}{4} = 27 \)
\[ k + 3 = 108 \implies k = 105 \]

**Case 2**: \( \frac{k + 9}{2} = 27 \)
\[ k + 9 = 54 \implies k = 45 \]

5. **Check the validity of \( k \)**:
- For \( k = 105 \):
- \( f(105) = 108 \)
- \( f(108) = 54 \)
- \( f(54) = 27 \) (valid)
- For \( k = 45 \):
- \( f(45) = 48 \)
- \( f(48) = 24 \)
- \( f(24) = 12 \) (not valid)

Thus, the only valid solution is \( k = 105 \).

### Final Answer:
The value of \( k \) is **105**.