The answer is B and this is how.

We have to see if n is odd.

St1: n can divisible by 3 and still be even. eg. 3,6,9,12.....
no enough

St2: 2n is divisible by twice as many number as by n. This is only possible if n is odd.

eg. if n=4; 2n=8 and they would both be divisible by only 2

if n=9 2n=18, now 2n is divisible by both 2 and 3 vs. only 3 with n.

Picking a few numbers and seeing a patttern is a good way to solve.

Conceptually this is what happens:

If n was even 2 would be one of its factor and so is 1 (1 is a factor of every integer)

If u mutliply by 2 each of the factors the factors will not double since 2*1 =2 is already a factor of n so 2n misses out on the double the factors part

So n cant be even


If n was odd then all its factors are odd.

When u multiply by 2 each odd factor of n u get a new even factor for 2n thereby doubling its factors (even*odd=EVEN)


Hope this helps too!

Regards,
CR

Excellent explanation cramya,
I couldnt have explained better.

B for me as well

rgds
-V

I just came across this question, and got the right OA.
However reading the explanation above (Cramya's in particular) I am a little confused.

If n = 3, then its factors are 1 and 3 (thus it has 2 factors). So 2n = 6, and its factors are 1,3 and 2. But this does not mean that 2n is "divisible by twice as many positive factors", as you have 3 factors now instead of 2.

So for me, the only answer to this question is when n = 1, as the only factor it has is 1, and therefore 2n = 2 has 2 factors.

Can someone please tell me how I have misinterpreted the question?
Thanks