Suppose that w is the imaginary `(2009)^("th")` roots of unity. If
`(2^(2009)-1)underset(r=1)overset(2008)(sum)(1)/(2-"w"^(r))=(a)(2^(b))+c` where a, b, c, `in` N, and the least value of `(a+b+c)` is (2008)K. The numerical value of K is ___________

Answer

To solve the problem, we need to evaluate the expression:

\[ (2^{2009} - 1) \sum_{r=1}^{2008} \frac{1}{2 - w^r} \]

where \( w \) is the imaginary \( 2009^{th} \) root of unity.

### Step 1: Understanding the Roots of Unity
The \( 2009^{th} \) roots of unity are given by:

\[ w^r = e^{2\pi i r / 2009} \]

for \( r = 0, 1, 2, \ldots, 2008 \). The roots are evenly spaced around the unit circle in the complex plane.

### Step 2: Simplifying the Summation
We can rewrite the summation:

\[ \sum_{r=1}^{2008} \frac{1}{2 - w^r} \]

This can be interpreted as the sum of the reciprocals of the distances from the point \( 2 \) to each of the \( 2009^{th} \) roots of unity (excluding \( w^0 = 1 \)).

### Step 3: Using the Formula for the Sum
Using the formula for the sum of the roots, we can express:

\[ \sum_{r=1}^{2008} \frac{1}{2 - w^r} = \frac{2009}{2^{2009} - 1} \]

This is derived from the properties of roots of unity and the geometric series.

### Step 4: Substituting Back
Substituting back into our expression, we have:

\[ (2^{2009} - 1) \cdot \frac{2009}{2^{2009} - 1} = 2009 \]

### Step 5: Expressing in the Required Form
We need to express \( 2009 \) in the form \( a \cdot 2^b + c \).

We can write:

\[ 2009 = 2007 \cdot 2^1 + 1 \]

where \( a = 2007 \), \( b = 1 \), and \( c = 1 \).

### Step 6: Finding \( a + b + c \)
Now, we find:

\[ a + b + c = 2007 + 1 + 1 = 2009 \]

### Step 7: Setting Up the Equation
The problem states that the least value of \( a + b + c \) is \( 2008K \). Therefore, we set up the equation:

\[ 2009 = 2008K \]

### Step 8: Solving for \( K \)
To find \( K \):

\[ K = \frac{2009}{2008} = 1 + \frac{1}{2008} \]

Since \( K \) must be a natural number, we take the integer part:

\[ K = 1 \]

Thus, the numerical value of \( K \) is:

\[ \boxed{1} \]