Integers: Prime Numbers

If $$p$$ is a prime number, which of the following must be true? >I. $$2p$$ is not a prime number. >II. $$p^2$$ is a prime number. >III. $$\frac{p}{2}$$ is not an integer.

Incorrect. [[Snippet]] Eliminate II since the square of an integer cannot be prime because it will have at least 3 factors—the number 1, the number itself ($$p$$), and its square ($$p^2$$). For example, if $$p=2$$, then $$p^2 = 4$$ is not prime. Thus, II does not *have* to be true. Eliminate III because 2 is a prime number and $$\frac{2}{2}= 1$$, and 1 is an integer. Thus, III does not *have* to be true.

Incorrect. [[Snippet]] You're right about I, but are you sure about II?

Incorrect. [[Snippet]] Did you forget that 2 is a prime number? In the equation $$\frac{2}{2} = 1$$, the number 1 is an integer. Thus, III does not *have* to be true.

Incorrect. [[Snippet]] The square of an integer cannot be prime because it will have at least 3 factors—the number 1, the number itself ($$p$$), and its square ($$p^2$$). Thus, II does not *have* to be true.

Correct. [[Snippet]] It's true that $$2p$$ can never be a prime number because it has 3 factors (i.e., 1, $$p$$, and $$2p$$). Prime numbers have exactly two factors. Thus, I *must* be true. Eliminate II since the square of an integer cannot be prime because it will have at least 3 factors—the number 1, the number itself ($$p$$), and its square ($$p^2$$). For example, if $$p=2$$, then $$p^2 = 4$$ is not prime. Thus, II does not *have* to be true. Eliminate III because 2 is a prime number and $$\frac{2}{2}= 1$$, and 1 is an integer. Thus, III does not *have* to be true. Therefore, only statement I is true.

I only

II only

III only

I and II only

II and III