Can you explain how option D) and E) also be an answer. What I did was:
12 = 2^2 and 3^1
so N^2 must content atleast 2^2 and 3^1. So, N must contain atleast 2^1 and 3^1 therefore N/6 satisfies the requirement.

OE


As 𝑁 is a positive integer and 𝑁^2 is divisible by 12, hence it can be written as

N^2/12=I (I is an integer)


𝑂𝑟, \(𝑁^2 = 12𝐼 = 2 × 2 × 3 × 𝐼\)
𝑂𝑟, \(𝑁 =\sqrt{ 2*2*3*I}=2 \sqrt{3*I}\)

Now, the fact important to notice here is under the root is I because the root must be also an integer to have N= integer

the minimum value of \(\sqrt{3*I}\) is I=3

therefore the minimum value of N is 6

D) 2N/3=2*6/3=12/3=4 It is an integer

E) 4N/3=6*4/3=8 It is an integer

As N^2 is divisible by 12.
assume N =12
Therefore N^2 is divisible by 12.
looking at the options put N =12
Answer = B,D,E

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.