`A_(1),A_(2)…A_(n)` are points on the line y = x lying in the positive quadrant such that `OA_(n)=nOA_(n-1), O` being the origin. If `OA_(1)=1` and the coordinates of `A_(n)` are `(2520sqrt(2), 2520sqrt(2))`, then n =

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to find the value of \( n \) given the conditions of the points \( A_1, A_2, \ldots, A_n \) on the line \( y = x \) in the positive quadrant.

### Step-by-Step Solution:

1. **Understanding the Points on the Line**:
The points \( A_1, A_2, \ldots, A_n \) lie on the line \( y = x \). Therefore, the coordinates of each point \( A_k \) can be represented as \( (x_k, x_k) \).

2. **Given Information**:
We know that \( OA_1 = 1 \) and the coordinates of \( A_n \) are \( (2520\sqrt{2}, 2520\sqrt{2}) \). The distance from the origin \( O \) to the point \( A_k \) is given by \( OA_k = k \cdot OA_{k-1} \).

3. **Calculating Distances**:
- For \( A_1 \):
\[ OA_1 = 1 \]
- For \( A_2 \):
\[ OA_2 = 2 \cdot OA_1 = 2 \cdot 1 = 2 \]
- For \( A_3 \):
\[ OA_3 = 3 \cdot OA_2 = 3 \cdot 2 = 6 \]
- For \( A_4 \):
\[ OA_4 = 4 \cdot OA_3 = 4 \cdot 6 = 24 \]
- Continuing this pattern, we find that:
\[ OA_n = n \cdot OA_{n-1} = n! \quad \text{(factorial of n)} \]

4. **Finding \( OA_n \)**:
Since the coordinates of \( A_n \) are \( (2520\sqrt{2}, 2520\sqrt{2}) \), the distance \( OA_n \) can also be calculated using the distance formula:
\[ OA_n = \sqrt{(2520\sqrt{2})^2 + (2520\sqrt{2})^2} = \sqrt{2 \cdot (2520\sqrt{2})^2} = \sqrt{2 \cdot 2 \cdot 2520^2} = 2520 \cdot 2 = 5040 \]

5. **Setting Up the Equation**:
From the previous steps, we have:
\[ n! = 5040 \]

6. **Finding \( n \)**:
We need to find \( n \) such that \( n! = 5040 \). We can calculate the factorials:
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)
- \( 6! = 720 \)
- \( 7! = 5040 \)

Thus, \( n = 7 \).

### Final Answer:
\[ n = 7 \]