Let `n in N, f(n)={(log_(8)n" if "log_(8)n " is integer"),(0" otherwise"):}`, then the valud of `sum_(n=1)^(2011) f(n)` is :

Options

A. 2011
B. `2011xx1006`
C. 6
D. `2^(2011)`

Answer

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

\[ f(n) = \begin{cases} \log_8 n & \text{if } \log_8 n \text{ is an integer} \\ 0 & \text{otherwise} \end{cases} \]

We want to find the value of \( \sum_{n=1}^{2011} f(n) \).

### Step 1: Determine when \( \log_8 n \) is an integer
The logarithm \( \log_8 n \) is an integer if and only if \( n \) can be expressed as \( n = 8^k \) for some integer \( k \). Since \( 8 = 2^3 \), we can rewrite this as:

\[ n = 2^{3k} \]

for \( k \in \mathbb{N} \).

### Step 2: Identify the values of \( n \) that satisfy this condition
We need to find the values of \( n \) that are of the form \( 2^{3k} \) and also less than or equal to 2011.

Calculating the maximum \( k \):
\[ 2^{3k} \leq 2011 \implies 3k \leq \log_2(2011) \implies k \leq \frac{\log_2(2011)}{3} \]

Calculating \( \log_2(2011) \):
Using a calculator or logarithm tables, we find:
\[ \log_2(2011) \approx 11.0 \implies k \leq \frac{11.0}{3} \approx 3.67 \]

Thus, \( k \) can take the integer values \( 0, 1, 2, 3 \).

### Step 3: Calculate the corresponding values of \( n \)
Now we calculate \( n \) for \( k = 0, 1, 2, 3 \):
- For \( k = 0 \): \( n = 2^{3 \cdot 0} = 1 \)
- For \( k = 1 \): \( n = 2^{3 \cdot 1} = 8 \)
- For \( k = 2 \): \( n = 2^{3 \cdot 2} = 64 \)
- For \( k = 3 \): \( n = 2^{3 \cdot 3} = 512 \)

### Step 4: Calculate \( f(n) \) for these values
Now we evaluate \( f(n) \) for each of these values:
- \( f(1) = \log_8(1) = 0 \)
- \( f(8) = \log_8(8) = 1 \)
- \( f(64) = \log_8(64) = 2 \)
- \( f(512) = \log_8(512) = 3 \)

### Step 5: Sum the values of \( f(n) \)
Now we sum these values:
\[ \sum_{n=1}^{2011} f(n) = f(1) + f(8) + f(64) + f(512) = 0 + 1 + 2 + 3 = 6 \]

### Final Answer
Thus, the value of \( \sum_{n=1}^{2011} f(n) \) is \( \boxed{6} \).