If `2^(2010)=a_(n)10^(n)+a_(n-1)10^(n-1)+………..+a_(2)10^(2)+a_(1)*10+a_(0)`, where `a_(i) in {0, 1, 2, ………, 9}` for all `i=0,1, 2,3, …………, n`, then `n=`

Options

A. 603
B. 604
C. 605
D. 606

Answer

To solve the problem, we need to find the value of \( n \) in the representation of the number \( 2^{2010} \) in the form of a polynomial in base 10. The steps are as follows:

### Step 1: Understand the representation
The number \( 2^{2010} \) can be expressed in the form:
\[ 2^{2010} = a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 \]
where \( a_i \) are the digits of the number in base 10.

### Step 2: Determine the number of digits
To find \( n \), we need to determine the number of digits in \( 2^{2010} \). The number of digits \( d \) in a number \( x \) can be found using the formula:
\[ d = \lfloor \log_{10} x \rfloor + 1 \]
Thus, we need to calculate:
\[ d = \lfloor \log_{10} (2^{2010}) \rfloor + 1 \]

### Step 3: Apply logarithmic properties
Using the properties of logarithms, we can simplify:
\[ \log_{10} (2^{2010}) = 2010 \cdot \log_{10} 2 \]
Now, we need to find \( \log_{10} 2 \). The approximate value is:
\[ \log_{10} 2 \approx 0.301 \]
So we can substitute this value into our equation:
\[ \log_{10} (2^{2010}) \approx 2010 \cdot 0.301 = 604.01 \]

### Step 4: Calculate the number of digits
Now we can find the number of digits:
\[ d = \lfloor 604.01 \rfloor + 1 = 604 + 1 = 605 \]

### Step 5: Find \( n \)
Since \( n \) is defined as the number of digits minus 1:
\[ n = d - 1 = 605 - 1 = 604 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{604} \]