If k is a positive integer, is k a prime number?

If k is a positive integer, is k a prime number?

Given: \(k=integer>0\). Question: \(k=prime\)?

A prime number is a positive integer with exactly two distinct divisors: 1 and itself.

(1) No integer between 2 and \(\sqrt{k}\) inclusive divides k evenly --> let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\). As given that \(k\) has no factor between 2 and \(\sqrt{k}\) inclusive, then both factors \(a\) and \(b\) must be more than \(\sqrt{k}\). But it's not possible, as the product of two positive integers more than \(\sqrt{k}\) will yield an integer more than \(k\) (\(ab>k\)). Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.

(2) No integers between 2 and \(\frac{k}{2}\) inclusive divides k evenly, and k is greater than 5 --> the same here : let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\) --> \(k=ab\geq{\frac{k^2}{4}}\) (as both \(a\) and \(b\) are more than or equal to \(\frac{k}{2}\), then their product \(ab\), which is \(k\), must be more than or equal to \(\frac{k}{2}*\frac{k}{2}\)) --> \(4k\geq{k^2}\) --> \(k(4-k)\geq{0}\). But this inequality cannot be true as \(4-k\) will be negative (as given \(k>5\)) and \(k\) is positive so \(k(4-k)\) must be negative not positive or zero. Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.

Answer: D.

P.S. The first statement is basically the way of checking whether some # is a prime:

Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers (primes) smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq \sqrt{n}\).
Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.

Hope it helps.