The first thing to do is narrow things down a little and say that m is a positive odd integer selected from the set of odd integers from 3 to 29.

I notice that both 3 and 29 are each divisible by only one prime number. Between them lie numbers like 15 and 21, which are divisible by more than one prime number.

I personally don't even want to think about this much and the list is pretty limited. So I am just going to list the odd numbers.

3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Most of them are either prime, and so divisible by only one prime number, or squares or a cube, 27, and so divisible by only one prime number. I glanced at the statements already and I see where this is headed.

The only ones divisible by more than one prime number are 15 and 21.

Statement 1 says that m is not divisible by 3. So 15 and 21 can both be eliminated. Since every other number in the list is divisible by only one prime number, we have our answer and Statement 1 is sufficient.

Statement 2 tells us that m is not 15, but m could still be 21. So we can't determine the answer to the question and Statement 2 is therefore insufficient.

So the correct answer is A.

A quick approach:

Since 2 < m < 30 and m is odd, m can have at most TWO unique prime factors. This is because 3 * 5 * 7 = 105, so if m has THREE unique prime factors, it will be greater than 30.

If m has exactly two unique prime factors, m must be a multiple of 3*5 or 3*7, as these are the only pairs of odd primes whose product is less than 30.

Given S1, m is not divisible by 3, so m cannot be a multiple of 3*5 or 3*7. Hence m cannot have two distinct odd prime factors, and it must only have ONE; SUFFICIENT.

Given S2, m could be of the form 3*7, or it could just be something else (like 23, or 3*3, or whatever); NOT SUFFICIENT.