The sum of first n odd numbers (i.e., `1 + 3 + 5 + 7 + ….+ 2n - 1)` is divisible by 11111 then the value of n is :

Options

A. a. 12345
B. b. 11111
C. c. can't be determined
D. d. none of these

Answer

To solve the problem, we need to find the value of \( n \) such that the sum of the first \( n \) odd numbers is divisible by 11111.

### Step-by-Step Solution:

1. **Understanding the Sum of Odd Numbers**:
The sum of the first \( n \) odd numbers can be expressed using the formula:
\[ S_n = 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]
This means the sum of the first \( n \) odd numbers is equal to \( n^2 \).

2. **Setting Up the Divisibility Condition**:
We need to find \( n \) such that:
\[ n^2 \text{ is divisible by } 11111 \]
This can be written mathematically as:
\[ n^2 \mod 11111 = 0 \]

3. **Finding the Prime Factorization of 11111**:
To understand the divisibility condition better, let's factor 11111.
Upon checking, we find that:
\[ 11111 = 41 \times 271 \]
This means for \( n^2 \) to be divisible by 11111, \( n \) must be divisible by both 41 and 271.

4. **Finding the Least Common Multiple**:
The least common multiple of 41 and 271 is:
\[ \text{lcm}(41, 271) = 41 \times 271 = 11111 \]
Therefore, \( n \) must be a multiple of \( \sqrt{11111} \).

5. **Calculating the Value of \( n \)**:
The smallest integer \( n \) that satisfies this condition is:
\[ n = 11111 \]

6. **Conclusion**:
Thus, the value of \( n \) such that the sum of the first \( n \) odd numbers is divisible by 11111 is:
\[ n = 11111 \]