Yes I think the answer has to be B. Statement II alone sufficient.

Given: Statement II is sufficient:

2n is divisible by twice as many positive integers as n.

you can just try some numbers here:

But remember what is the question ?

Question is, Is n odd ?

Let try an odd number,

n = 3(Its also a prime number) ,  then factors are 1 and 3

2n = 6, then factors are 1, 2, 3, and 6

Yes as per the given statement it has exactly twice the number of factors.

Even number,

n = 4, then factors are 1,2 and 4

2n = 8, then factors are 1,2,4 and 8, so there are not twice as many factors are here, so we can’t take a even number here.

So, then try even prime number which is 2.

2n = 4, then factors are 1,2 and 4, so there are only factors, so this doesn’t fit.

lets try an odd number which is not prime,

n= 9,  then factors are 1,3 and 9

2n = 18, then factors are 1, 2, 3,6,9 and 18

Yes as per the given statement it has exactly twice the number of factors.

So, it has to be an odd number.

So sufficient.

Technically, the math here is, If “n” is an odd number, then 2*n will have twice as many factors as n, the converse also true. Because 2 is an even number.

Hope this helps.

Is OA B?