If N is the sum of first 13,986 prime numbers, then N is always divisible by

Options

A. 6
B. 4
C. 8
D. None of these

Answer

To determine whether the sum \( N \) of the first 13,986 prime numbers is divisible by 6, 4, 8, or none of these, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the first prime number**: The first prime number is 2. It is also the only even prime number.

2. **Separate the first prime number from the sum**:
\[ N = 2 + \text{(sum of the next 13,985 prime numbers)} \]

3. **Consider the remaining prime numbers**: The remaining 13,985 prime numbers are all odd (since all prime numbers greater than 2 are odd).

4. **Sum of odd numbers**: The sum of an odd number of odd numbers is odd. Therefore, the sum of these 13,985 odd prime numbers is odd.

5. **Combine the sums**:
\[ N = 2 + \text{(odd number)} \]
Since 2 is even and the sum of odd numbers is odd, the result \( N \) will be:
\[ \text{even} + \text{odd} = \text{odd} \]

6. **Determine divisibility by even numbers**: Since \( N \) is odd, it cannot be divisible by any even number. The options provided (6, 4, and 8) are all even.

7. **Conclusion**: Since \( N \) is odd, it is not divisible by 6, 4, or 8. Therefore, the correct answer is:
\[ \text{None of these} \]

### Final Answer:
N is always divisible by **none of these**.

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