Divisibility problem in FOM I recording

[nuwalt]

Stacey,

I was reviewing the FOM I recording and came across a questions which I thought I did it right, however, when the solution was presented, I got it wrong. Don't know why. Here is the question:

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

I. 3
II. 18
III. 27

Answer choices:

(A) I only
(B) I and II only
(C) II and III only
(D) I and III only
(E) I, II, and III

Initially, I put the answer choice (E) because of the following:

Since x is divisible by 6 and 9, if you take prime factors of 6 is 3, 2. The prime factors of 9 is 3, 3. So x has the following prime factors: 2, 3, 3, 3. It could also have other factors which is a quesiton but atleast its telling us what we have. Now going to the main question...

Yes, x must be a multiple of 3 since x is divisible by 6 and 6 is divisible by 3. Also, yes x must be a multiple of 18 since x has prime factors that if we multiply 2, 3, 3, it gives us 18.

I did the same thing for 27, since 27's prime factors are 3, 3, 3. These three 3's appear in the prime factors of x also, so why x is not a multiple of 27.

The correct answer according to the FOM record is (B), i.e. only x is a multiple of 3 and 18....WHY NOT 27????

Please help. Thanks.

[MannyRosas]

If X is divisible by 6 and 9.

Since x is divisible by 6 and 9, if you take prime factors of 6 is 3, 2. The prime factors of 9 is 3, 3. So x has the following prime factors: 2, 3, 3, 3.

nuwalt,
you were right at the beginning, but not at the end.

prime factors of 6 are 3, 2
prime factors of 9 are 3, 3
prime factors of X are 2, 3, 3 and could have other factors (but we don't care because they were not given). Notice that only include 2 3s because the the 3 from 6 is already included in the 3s from 9.

Hence, X must be a multimple of any combinations of
2, 3, 3 = 2, 3, 6, 9, 18

(B) I and II only

hope it helps

[tim]

Thanks Manny. Nuwalt, let us know if you have any further questions about this one..