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Re: If m is divisible by 3, how many prime factors does m have? [#permalink]
vivektripathi wrote:
If m is divisible by 3, how many prime factors does m have?
1). m/3 is divisible by 3.
2). m/3 has two different prime factors.


(1) m is divisible by 3 and m/3 is divisible by 3, this means m is divisible by 9

possible m values = {9,18,27...} 9 has 2 PF (3 X3) , 18 has 3 PF ( 2X3X3) Insuff

(2) m is divisible by 3 { 3,6,9,12,15,18...} and m/3 has two diff prime factors .

Does it mean that m/3 has only 2 different prime factors such as ( 2 and 3) or can it be (2 , 2 and 3) still be two different prime factors. It seems the later but confused.
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Re: If m is divisible by 3, how many prime factors does m have? [#permalink]
vivektripathi wrote:
If m is divisible by 3, how many prime factors does m have?
1). m/3 is divisible by 3.
2). m/3 has two different prime factors.

1) is insufficient since clearly m can be 21[3],27[3],54[4] etc
2)m/3 has 2 diff prime factors means that 3,2 or 2,7 means m has 3 prime factors IMO B
even im confused as to 2 prime factors(which are diff) of or set of two diff prime factors

whats the source of this question?can we expect such Q in GMAT :shock:
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Re: If m is divisible by 3, how many prime factors does m have? [#permalink]
Why could not be C?

As, m/3 is divisible by 3. So one prime factor is 3.

And, m/3 has two different prime number so one is 3 and second could be any thing..

In that case, m has only two prime factors: 3 and any other.

So, Answer is C.
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Re: If m is divisible by 3, how many prime factors does m have? [#permalink]
vivektripathi wrote:
If m is divisible by 3, how many prime factors does m have?
1). m/3 is divisible by 3.
2). m/3 has two different prime factors.


from tem

m= 3x

from 1

3x/3 = intiger x ( m at least have 3^2 and some other unkonwn factors

from 2

x includes 2 different prime facotrs

thus 3x has 3 different prime factors .........thus m has 3 prime factors ........B is my answer
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Re: If m is divisible by 3, how many prime factors does m have? [#permalink]
Twoone wrote:
Why could not be C?

As, m/3 is divisible by 3. So one prime factor is 3.

And, m/3 has two different prime number so one is 3 and second could be any thing..

In that case, m has only two prime factors: 3 and any other.

So, Answer is C.



You are right. It should be C. I simply overlooked this.

From stmt1 and stmt2: n/3 could be 6,12,15,18..... and hence n will always have 3 and one more prime factor.



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