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If a positive integer q is divisible by both 3 and 11, then q must als

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If a positive integer q is divisible by both 3 and 11, then q must als  [#permalink]

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New post 03 Mar 2011, 20:56
3
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

74% (00:45) correct 26% (00:31) wrong based on 56 sessions

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If a positive integer q is divisible by both 3 and 11, then q must also be divisible by which of the following?

I. 14
II. 33
III. 66

A. I only
B. II only
C. III only
D. I and II
E. II and III


Hello, I have been using The Princeton Review's Math Workout for the Gmat, 3rd Edition, and encountered a question that I don't fully understand.

Practice Set, pg. 58, #5

I chose E, but the correct answer is B. The explaination given is as follows:

Numbers that are divisble by 3 and 11 include 33, 66, 99, 132, and so forth. You can eliminate I and III because not all these numbers are divisible by 14 or 66. All these numbers are divisble by 33, so choose B.

I still don't understand why 66 is eliminated. I would appreciate it if someone could explain this to me; I am ridiculously slow-witted with math, but what can ya do?

Thanks!

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Re: If a positive integer q is divisible by both 3 and 11, then q must als  [#permalink]

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New post 03 Mar 2011, 21:08
66 is eliminated because 33 is not divisible by 66. it is just divisible by 3 and 11.
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Re: If a positive integer q is divisible by both 3 and 11, then q must als  [#permalink]

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New post 03 Mar 2011, 21:49
But what if q is 66? if q is 66 its divisible by 3, 11, 33 and itself 66. No?
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Re: If a positive integer q is divisible by both 3 and 11, then q must als  [#permalink]

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New post 04 Mar 2011, 00:03
1
This hinges on the wording of the question. The question asks what q *must* be divisible by. It certainly is *possible* that q=66, say, and that q is divisible by 66, but it's also possible that q = 33, in which case q is not divisible by 66. So q does not *need* to be divisible by 66 here, which is why we can rule out III. I'd add that it's also *possible* for q to be divisible by 14 (it might be that q = 3*11*14 = 462, for example) but q certainly does not need to be divisible by 14. However, if q is divisible by 3 and 11, then q will always be divisible by the LCM of 3 and 11, which is 33, so q must be divisible by 33.
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Re: If a positive integer q is divisible by both 3 and 11, then q must als  [#permalink]

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New post 04 Mar 2011, 02:17
IanStewart wrote:
This hinges on the wording of the question. The question asks what q *must* be divisible by. It certainly is *possible* that q=66, say, and that q is divisible by 66, but it's also possible that q = 33, in which case q is not divisible by 66. So q does not *need* to be divisible by 66 here, which is why we can rule out III. I'd add that it's also *possible* for q to be divisible by 14 (it might be that q = 3*11*14 = 462, for example) but q certainly does not need to be divisible by 14. However, if q is divisible by 3 and 11, then q will always be divisible by the LCM of 3 and 11, which is 33, so q must be divisible by 33.


Well fancy that, thanks a lot IanStewart! I get it!=D
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Re: If a positive integer q is divisible by both 3 and 11, then q must als  [#permalink]

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Re: If a positive integer q is divisible by both 3 and 11, then q must als &nbs [#permalink] 22 Sep 2018, 05:14
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