For some natural number N, the number of positive integral 'x' satisfying the equation, `1! +2!+ 3!+.............(x!)=(N)^2` is

Options

A. one
B. two
C. three
D. None of these

Answer

To solve the equation \(1! + 2! + 3! + \ldots + x! = N^2\) for positive integral values of \(x\), we will analyze the left-hand side and determine when it equals a perfect square.

### Step-by-Step Solution:

1. **Calculate the sum of factorials for small values of \(x\)**:
- For \(x = 1\):
\[ 1! = 1 \]
This is \(1^2\), so \(N = 1\).

- For \(x = 2\):
\[ 1! + 2! = 1 + 2 = 3 \]
This is not a perfect square.

- For \(x = 3\):
\[ 1! + 2! + 3! = 1 + 2 + 6 = 9 \]
This is \(3^2\), so \(N = 3\).

- For \(x = 4\):
\[ 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 \]
This is not a perfect square.

- For \(x = 5\):
\[ 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 \]
This is not a perfect square.

- For \(x = 6\):
\[ 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 \]
This is not a perfect square.

- For \(x = 7\):
\[ 1! + 2! + 3! + 4! + 5! + 6! + 7! = 873 + 5040 = 5913 \]
This is not a perfect square.

2. **Identify the pattern**:
- We found that the sums for \(x = 1\) and \(x = 3\) yield perfect squares, while for \(x = 2, 4, 5, 6, 7\) they do not.
- As \(x\) increases, the factorial values grow rapidly, making it less likely for the sum to be a perfect square.

3. **Conclusion**:
- The only positive integral values of \(x\) that satisfy the equation \(1! + 2! + 3! + \ldots + x! = N^2\) are \(x = 1\) and \(x = 3\).
- Therefore, the number of positive integral \(x\) satisfying the equation is **2**.

### Final Answer:
The number of positive integral \(x\) satisfying the equation is **2**.