On the plane are given $ k\plus{}n$  distinct lines , where $ k>1$ is integer and $ n$ is integer as well.Any three of these lines do not pass through the

 same point . Among these lines exactly  $ k$ are parallel and all the other $ n$ lines intersect each other.All $ k\plus{}n$ lines define on the plane a partition 

of triangular , polygonic or not bounded regions. Two regions are colled different, if the have not common points 

or if they have common points only on their boundary.A regions is called ''good'' if it contained in a zone between two parallel lines .

If in a such given configuration the minimum number of ''good'' regionrs is $ 176$ and the maximum number of these regions is $ 221$, find $ k$ and $ n$.


Babis
1. **Define the problem and variables:**
   - We have \( k + n \) distinct lines on a plane.
   - \( k \) lines are parallel (\( k > 1 \)).
   - \( n \) lines intersect each other and are not parallel to the \( k \) lines.
   - The minimum number of "good" regions is 176.
   - The maximum number of "good" regions is 221.

2. **Minimum number of good regions:**
   - Good regions are those contained in the strip between the \( k \) parallel lines.
   - The minimum number of good regions occurs when all intersections of the \( n \) lines are outside the strip.
   - Each of the \( n \) lines divides the strip into \( k - 1 \) regions.
   - Therefore, the minimum number of good regions is:
     \[
     f_{\text{min}} = (n + 1)(k - 1)
     \]
   - Given \( f_{\text{min}} = 176 \):
     \[
     (n + 1)(k - 1) = 176
     \]

3. **Maximum number of good regions:**
   - The maximum number of good regions occurs when all intersections of the \( n \) lines are inside the strip.
   - Each new line added intersects all previous lines, creating additional segments.
   - The number of segments added by the \( i \)-th line is \( k + i - 2 \).
   - Therefore, the maximum number of good regions is:
     \[
     f_{\text{max}} = 2(k - 1) + \sum_{i=2}^n (k + i - 2)
     \]
   - Simplifying the sum:
     \[
     \sum_{i=2}^n (k + i - 2) = (n-1)k + \sum_{i=2}^n (i - 2) = (n-1)k + \sum_{i=0}^{n-2} i = (n-1)k + \frac{(n-2)(n-1)}{2}
     \]
   - Therefore:
     \[
     f_{\text{max}} = 2(k - 1) + (n-1)k + \frac{(n-2)(n-1)}{2}
     \]
   - Given \( f_{\text{max}} = 221 \):
     \[
     2(k - 1) + (n-1)k + \frac{(n-2)(n-1)}{2} = 221
     \]

4. **Solve the equations:**
   - From the minimum number of good regions:
     \[
     (n + 1)(k - 1) = 176
     \]
   - From the maximum number of good regions:
     \[
     2(k - 1) + (n-1)k + \frac{(n-2)(n-1)}{2} = 221
     \]
   - Subtract the minimum from the maximum:
     \[
     221 - 176 = 45
     \]
     \[
     \frac{n^2 - n}{2} = 45
     \]
     \[
     n^2 - n - 90 = 0
     \]
     \[
     n = \frac{1 \pm \sqrt{1 + 360}}{2} = \frac{1 \pm 19}{2}
     \]
     \[
     n = 10 \quad \text{(since } n \text{ must be positive)}
     \]

5. **Find \( k \):**
   - Substitute \( n = 10 \) into the minimum equation:
     \[
     (10 + 1)(k - 1) = 176
     \]
     \[
     11(k - 1) = 176
     \]
     \[
     k - 1 = 16
     \]
     \[
     k = 17
     \]

The final answer is \( \boxed{ k = 17 } \) and \( n = 10 \).