Solution:

We need to find, whether \(N^3/4\) is an integer or not?

Now, \(N^3/4\) is an integer only when \(N\) is even number.

Thus, we only need to find, whether \(N\) is an even number or not?

Statement 1:"\(N^2+3\) is a prime number.”

Since \(N\) is a positive integer, therefore, \(N^2+3\) is always greater than \(3\).

We know, all Prime numbers greater than \(3\) are odd.

Thus,

\(N^2+3\) = \(odd\)

\(N^2\)+ \(odd=odd\)

\(N^2\)= \(even\)

Therefore, \(N^2+3\) is a prime number when \(N\) is an even number.

Thus, \(N^3/4\) is an integer.

Therefore, Statement 1 alone is sufficient to answer the question.

Statement 2:“\(N\) is the number of odd factors of \(6\).”

To calculate odd factors of a number, we do not consider the even prime number. Therefore,

Odd factors of \(6\)= power of \(3+1\)

Odd factors of\(6\)= \(1+1\)

Odd factors of \(6=2\)

Thus, \(N=2\).

Therefore, \(N^3/4\) is an integer.

Therefore, Statement 2 alone is sufficient to answer the question.

Therefore, we can find the answer by EACH of the statement alone.

Answer: Option D

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