1. Let n be a positive integer such that 1 ≤ n ≤ 1000 . Let Mn be the number of integers in the set Xn ={√4n+1,√4n+2...................√4n+1000}. let a = max{Mn : 1 ≤ n ≤ 1000}, and b = min{Mn : 1 ≤ n ≤ 1000}.
find a-b

To solve this problem, we need to analyze the set \( X_n = \{\sqrt{4n+1}, \sqrt{4n+2}, \ldots, \sqrt{4n+1000}\} \) for \( n \) ranging from 1 to 1000. The goal is to determine the values of \( M_n \), which represents the number of integers in the set \( X_n \), and then find \( a = \max(M_n) \) and \( b = \min(M_n) \), and finally compute \( a - b \).

### Step-by-Step Solution

1. **Understanding \( M_n \):**

   The set \( X_n \) contains square roots of consecutive integers from \( 4n+1 \) to \( 4n+1000 \). To find \( M_n \), we need to count how many of these square roots are integers.

   A square root \( \sqrt{k} \) is an integer if and only if \( k \) is a perfect square. Therefore, \( M_n \) is the number of perfect squares within the interval \([4n+1, 4n+1000]\).

2. **Determine the Range of Perfect Squares:**

   Let \( A = 4n+1 \) and \( B = 4n+1000 \). We want to find the integers \( m \) such that \( m^2 \) lies in the range \([A, B]\).

   This translates to:

   \[
   \sqrt{A} \leq m \leq \sqrt{B}
   \]

   Plugging in the values of \( A \) and \( B \), we get:

   \[
   \sqrt{4n+1} \leq m \leq \sqrt{4n+1000}
   \]

   The number of integers \( m \) satisfying this inequality is:

   \[
   M_n = \lfloor \sqrt{4n+1000} \rfloor - \lceil \sqrt{4n+1} \rceil + 1
   \]

3. **Maximizing and Minimizing \( M_n \):**

   To find \( a = \max(M_n) \) and \( b = \min(M_n) \), we need to understand how \( M_n \) varies with \( n \). 

   The expressions \( \sqrt{4n+1} \) and \( \sqrt{4n+1000} \) are both increasing functions of \( n \). Consequently, \( M_n \) could vary as \( n \) changes from 1 to 1000.

4. **Approximate \( M_n \) for Extremes:**

   - When \( n = 1 \):
   
     \[
     A = 4 \times 1 + 1 = 5, \quad B = 4 \times 1 + 1000 = 1004
     \]

     \[
     \sqrt{A} \approx \sqrt{5} \approx 2.236, \quad \sqrt{B} \approx \sqrt{1004} \approx 31.654
     \]

     So,

     \[
     M_1 = \lfloor 31.654 \rfloor - \lceil 2.236 \rceil + 1 = 31 - 3 + 1 = 29
     \]

   - When \( n = 1000 \):

     \[
     A = 4 \times 1000 + 1 = 4001, \quad B = 4 \times 1000 + 1000 = 5000
     \]

     \[
     \sqrt{A} \approx \sqrt{4001} \approx 63.25, \quad \sqrt{B} \approx \sqrt{5000} \approx 70.71
     \]

     So,

     \[
     M_{1000} = \lfloor 70.71 \rfloor - \lceil 63.25 \rceil + 1 = 70 - 64 + 1 = 7
     \]

   From these computations, we observe that \( M_1 = 29 \) and \( M_{1000} = 7 \).

5. **Find \( a \) and \( b \):**

   Given the range from \( n = 1 \) to \( n = 1000 \), we have:

   \[
   a = \max(M_n) = 29, \quad b = \min(M_n) = 7
   \]

6. **Compute \( a - b \):**

   \[
   a - b = 29 - 7 = 22
   \]

### Final Answer

\[
\boxed{22}
\]

Would you like more details or have any questions? Here are five related questions to explore:

1. How does the function \( M_n \) behave as \( n \) increases?
2. What is the impact of the range \( [4n+1, 4n+1000] \) on the values of \( M_n \)?
3. How would the problem change if the range was altered from 1000 to another number?
4. Can you find a general formula for \( M_n \) for any range?
5. How does the ceiling and floor function affect the calculation of \( M_n \)?

**Tip:** Understanding the behavior of square roots and perfect squares is crucial for solving problems involving ranges of numbers.

Let n be a positive integer such that 1 ≤ n ≤ 1000. Let Mn be the number of integers in the set Xn ={√4n+1,√4n+2...................√4n+1000}. Let a = max{Mn : 1 ≤ n ≤ 1000}, and b = min{Mn : 1 ≤ n ≤ 1000}. Find a-b.

Explore how to find the maximum and minimum values of Mn, the number of integers in a set of square roots for 1 ≤ n ≤ 1000. This detailed solution includes key formulas and theorems for working with square roots and perfect squares, suitable for Grades 9-12.

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