Hello,

I am having trouble understanding this question, and while the book does a great job explaining things, I seem to be unable to grasp the concept:

If x is divisible by 6 and 9, then x must be a multiple of

I. 3
II. 18
III. 27

I only
I and II only
II and III only
I and III only
I, II, and III
The prime factors of 6 are 2 and 3. So x must be divisible by at least one 2 and at least one 3.

The prime factors of 9 are 3 and 3. So x must be divisible by at least two 3's.

Putting these two pieces of evidence together, we conclude that the prime factors of x include at least one 2 and at least two 3's. Thus, x must be divisible by 2×3×3, which equals 18. Moreover, x also has to be divisible by all the factors of 18, including 3. However, there is no need for x to be divisible by 27 (=3×3×3), since we don't know whether x contains a third 3.

The correct answer is B.

If X is divisible by both, 6 and 9, would X not have one 2 and three 3's (6=2x3 & 9=3x3)?

I chose I II and III, and at first it did not make sense, but when I did my amateur version of the math, 27 seemed to fit in as well.

Thanks ahead of time,

Milan

Milan,
You nicely worked out that x must be a multiple of 18. x definitely contains one 2 and two 3s, which results in a product of 18. x may have other factors, but we know it has those. Yes to statement II.

Also, if x is a multiple of 18, it must also be a multiple of each of 18's factors. Thus, since 3 is a factor of 18, x must also be a multiple of 3. Yes to statement I.

However, does x have to be a multiple of 27? No. Based on the information we have, x could be 18, which is not a multiple of 27. If we knew that we have a third 3 then we would know x is a multiple of 27, but since we don't have that knowledge we cannot definitively conclude that statement III is true.

I hope this answers your questions.

_________________
Jamie Nelson
ManhattanGMAT Instructor

Jaime,

Thank you very much for your reply!

I am having a great deal of confusion in regards to the number of three's in this question. Please bear with me as I try to comprehend the info.

My understanding is that N would contain one 2 and three 3's: 2X3=6 and 3X3=9.

27 also has three 3's (3^3). In order for for X to be a multiple of 27, it would have to have all of the same prime factors as 27, which (under my logic) it does.

You made a great point about X possibly being 18, which proves that I am incorrect. However I am not sure where my logic was wrong.

Regards,

Milan

_________________
Jamie Nelson
ManhattanGMAT Instructor