If `N = (12345)^(2) + 12345 + 12346`, then what is the value of `sqrt(N)` ?

Options

A. A) 12346
B. B) 12345
C. C) 12344
D. D) 12347

Answer

To solve the problem, we need to find the value of \( \sqrt{N} \) where \( N = (12345)^2 + 12345 + 12346 \).

### Step-by-Step Solution:

1. **Define the variable**: Let \( a = 12345 \). Then we can rewrite \( N \) as:
\[ N = a^2 + a + (a + 1) \]

2. **Simplify the expression for \( N \)**: Combine the terms in \( N \):
\[ N = a^2 + a + a + 1 = a^2 + 2a + 1 \]

3. **Recognize the perfect square**: Notice that \( a^2 + 2a + 1 \) can be factored:
\[ N = (a + 1)^2 \]

4. **Substitute back the value of \( a \)**: Since \( a = 12345 \), we have:
\[ N = (12345 + 1)^2 = (12346)^2 \]

5. **Find the square root of \( N \)**: Now, we can find \( \sqrt{N} \):
\[ \sqrt{N} = \sqrt{(12346)^2} = 12346 \]

### Final Answer:
Thus, the value of \( \sqrt{N} \) is \( 12346 \).

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