If m is divisible by 3, how many prime factors does m have?

If m is divisible by 3, how many prime factors does m have?

(1) \(\frac{m}{3}\) is divisible by 3 --> \(\frac{m}{3}=3k\) --> \(m=3^2*k\) --> \(m\) has at least one prime 3, but it can have more than one, in case \(k\) has some number of other primes. Not sufficient

2) \(\frac{m}{3}\) has two different prime factors --> first of all 3 is a factor of \(m\), so 3 is one of the primes of \(m\) for sure.

Now, if power of 3 in \(m\) is more than or equal to 2 then \(m\) will have have only two prime factors: 3 and one other, example: \(m=18\), (as in \(\frac{m}{3}\) one 3 will be reduced, at least one more 3 will be left, plus one other, to make the # of different factors of \(\frac{m}{3}\) equal to two. Thus \(m\) will have 3 and some other prime as a prime factors).

But if \(m\) has 3 in power of one then \(m\) will have 3 prime factors: 3 and two others, example \(m=30\) (one 3 will be reduced in \(m\) and \(\frac{m}{3}\) will have some other two prime factors, which naturally will be the primes of \(m\) as well). Not sufficient.

(1)+(2) From (1) \(3^2\) is a factor of \(m\), thus from (2) \(m\) has only two distinct prime factors: 3 and one other. Sufficient.

Answer: C.