prep 21，没有讨论过，看不懂题阿~~

It's generally more useful if you tell us what you struggled with so that we can target our answer to your needs.

n is an integer
is n odd?
yes/no question, so I will try to prove it wrong (that is, get a yes and a no based upon the statements)

(1) n/3
n could be 6 (that is divisible by 3). Is n odd? No
n could be 9 (that is divisible by 3). Is n odd? Yes
Elim A and D

(2) 2n has twice as many factors as n
n could be 1, which has one factor; 2n would be 2, which has two factors; is n odd? Yes
n could be 2, which has two factors; 2n would be 4, which has three factors. Oops, can't use this combo of numbers (has to make statement 2 true, and this combo doesn't)

What's going on here?
general rule: 2n will be divisible by 2 and also by whatever number 2n is.
If I make n an even number, even numbers are already divisible by 2. So 2n will only be divisible by one new number, equal to 2n. That is, I add only one new factor for 2n.
Any even number, by definition, has at least two factors - 1 and 2. So I would need to add at least two more factors to double the number of factors. But I can't - the setup of statement 2 only allows me to add one new factor if n is even. So I can never make statement 2 true using an even number for n.

Sufficient. Answer is B.