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. 12345
B. 11111
C. can't be determined
D. None of these

Answer

To solve the problem of finding the value of \( n \) such that the sum of the first \( n \) odd numbers is divisible by 11111, we can follow these steps:

### Step 1: Understand the formula for the sum of the first \( n \) odd numbers
The sum of the first \( n \) odd numbers can be expressed as:
\[ S_n = 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]
This means that the sum of the first \( n \) odd numbers is equal to \( n^2 \).

### Step 2: Set up the divisibility condition
We need to find \( n \) such that:
\[ n^2 \text{ is divisible by } 11111 \]
This can be written mathematically as:
\[ n^2 \equiv 0 \mod 11111 \]

### Step 3: Factor 11111
To solve this, we first factor 11111. We can check for divisibility by small prime numbers:
- 11111 is not divisible by 2 (it's odd).
- The sum of the digits \( 1 + 1 + 1 + 1 + 1 = 5 \) is not divisible by 3.
- The last digit is not 0 or 5, so it's not divisible by 5.
- Continuing this process, we find that:
\[ 11111 = 41 \times 271 \]

### Step 4: Determine the conditions for \( n \)
For \( n^2 \) to be divisible by \( 11111 \), \( n \) must be divisible by both factors of 11111. Thus, we need:
\[ n \equiv 0 \mod 41 \quad \text{and} \quad n \equiv 0 \mod 271 \]

### Step 5: Find the least common multiple
The least common multiple (LCM) of 41 and 271 is:
\[ \text{lcm}(41, 271) = 41 \times 271 = 11111 \]
Thus, \( n \) must be a multiple of 11111.

### Step 6: Conclusion
The smallest positive integer \( n \) that satisfies the condition is:
\[ n = 11111 \]

### Final Answer
The value of \( n \) is \( 11111 \).

---