Let n be a positive integer such that `log_2 log _2 log_2 log_2 log_2 (n) lt 0 lt log_2 log_2 log_2(n)`.
Let l be the number of digits in the binary expansion of n. Then the minimum and the maximum possible values of l are

Options

A. 5 and 16
B. 5 and 17
C. 4 and 16
D. 4 and 17

Answer

To solve the problem, we need to analyze the given inequalities involving logarithms.

### Step-by-Step Solution:

1. **Understanding the Inequalities**:
We have the inequalities:
\[ \log_2(\log_2(\log_2(\log_2(\log_2(n))))) < 0 < \log_2(\log_2(\log_2(n))) \]

2. **Analyzing the Left Inequality**:
From the left inequality, we can deduce:
\[ \log_2(\log_2(\log_2(\log_2(\log_2(n))))) < 0 \]
This implies:
\[ \log_2(\log_2(\log_2(\log_2(n)))) < 1 \]
Therefore:
\[ \log_2(\log_2(\log_2(n))) < 2 \]
Which leads to:
\[ \log_2(\log_2(n)) < 4 \]
Hence:
\[ \log_2(n) < 16 \]
This implies:
\[ n < 2^{16} \]

3. **Analyzing the Right Inequality**:
From the right inequality, we have:
\[ 0 < \log_2(\log_2(\log_2(n))) \]
This implies:
\[ \log_2(\log_2(n)) > 1 \]
Therefore:
\[ \log_2(n) > 2 \]
Which leads to:
\[ n > 2^2 = 4 \]

4. **Combining the Results**:
From the two inequalities we derived:
\[ 4 < n < 2^{16} \]

5. **Finding the Number of Digits in Binary Expansion**:
The number of digits \( l \) in the binary expansion of \( n \) can be found using:
\[ l = \lfloor \log_2(n) \rfloor + 1 \]
From \( n > 4 \), we have:
\[ \log_2(n) > 2 \quad \Rightarrow \quad l \geq 2 + 1 = 3 \]
From \( n < 2^{16} \), we have:
\[ \log_2(n) < 16 \quad \Rightarrow \quad l < 16 + 1 = 17 \]

6. **Final Values**:
Therefore, the minimum and maximum possible values of \( l \) are:
\[ \text{Minimum } l = 3, \quad \text{Maximum } l = 16 \]

### Conclusion:
The minimum and maximum possible values of \( l \) are \( 3 \) and \( 16 \), respectively.