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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]

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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, 21:17, edited 1 time in total.
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New post 08 Jan 2011, 21:14
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]

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

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

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

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New post 26 Jul 2016, 05:10
dimitri92 wrote:
\(q^2-5=( q^2-1) +4\)
= ( q+1) (q-1) +4



ummmm.. shouldn't \(q^2-5=( q^2-1)\)-4 ?????
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Re: For some integer q, q^2 - 5 is divisible by all of the follo [#permalink]

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New post 26 Jul 2016, 23:32
LogicGuru1 wrote:
dimitri92 wrote:
\(q^2-5=( q^2-1) +4\)
= ( q+1) (q-1) +4



ummmm.. shouldn't \(q^2-5=( q^2-1)\)-4 ?????


Yes, it should be.
Check: for-some-integer-q-q-2-5-is-divisible-by-all-of-the-follo-94414-20.html#p848986
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Re: For some integer q, q^2 - 5 is divisible by all of the follo   [#permalink] 26 Jul 2016, 23:32

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

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