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# For some integer q, q^2 - 5 is divisible by all of the follo

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Re: q^2 - 5 [#permalink]  08 Jan 2011, 20:10
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dimitri92 wrote:
For some integer q, q^2 - 5 is divisible by all of the following EXCEPT
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

The way I would approach this question:

So q^2 - 5 is divisible by all of the following except:
29, 31, 41 - big prime numbers, don't know any divisibility rules for these, forget them for the time being.. 38 = 19*2. (q^2 - 5) can be divisible by 2 (e.g. when q^2 ends with a 5, q^2 - 5 ends with a 0). As for 19, again a big prime number. Leave it for the time being.

(If the question is anywhere close to an actual GMAT question, they will not expect you to do many calculations with 29, 31, 41 etc. I see these big prime numbers and am quite convinced that they are just a smokescreen.Try and focus on what they could ask you like divisibility by 2, 3 etc. )

As for 30, q^2 - 5 is divisible by 10 (using the logic shown above). What about 3?
$$q^2 - 5 = q^2 - 1 - 4 = (q - 1)(q + 1) - 4$$
In any 3 consecutive numbers, (e.g. $$(q - 1), q, (q + 1)$$), one and only one number will be divisible by 3.
If either (q - 1) or (q + 1) is divisible by 3, (q - 1)(q + 1) is divisible by 3, which means $$(q - 1)(q + 1) - 4$$ cannot be divisible by 3. If q is divisible by 3, then q^2 will be divisible by 3 and q^2 - 5[/m] will not be divisible by 3.
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Last edited by VeritasPrepKarishma on 08 Jan 2011, 20:17, edited 1 time in total.
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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5464
Location: Pune, India
Followers: 1338

Kudos [?]: 6802 [0], given: 177

Re: q^2 - 5 [#permalink]  08 Jan 2011, 20:14
Expert's post
dimitri92: Didn't see your response since it was on page 2. But yes, that is exactly how I would think about it too. (Thought I think there is a small typo. You have a '+4' rather than a '-4')
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Re: For some integer q, q^2 - 5 is divisible by all of the follo [#permalink]  03 Oct 2013, 19:32
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Re: For some integer q, q^2 - 5 is divisible by all of the follo [#permalink]  31 Oct 2014, 21:37
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: For some integer q, q^2 - 5 is divisible by all of the follo   [#permalink] 31 Oct 2014, 21:37

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