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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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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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04 Sep 2017, 17:18

Bunuel wrote:

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

Hint: q^2-5 (q is an integer) is never multiple of 3 (try to prove this), hence 30 is out.

Answer: B.

Taking lead from Bunuel's post...

A) 29 - Prime B) 30 - 2 * 3 * 5 C) 31 - Prime D) 38 - 2 * 19 E) 41 - Prime

Let's start with simple numbers.

1) 2

Remainder when 5/2 is 1. And, perfect squares such as 81 produce remainder 1 when divided by 2. Thus, the overall remainder (1-1) is 0. 2 might divide. Park aside.

2) 3

Remainder when 5/3 is 2. The perfect squares when divided by 3, produce either 0 or 1 remainder. Case 1) When the remainder is 0 The remainder value of the expression is 0-2 = -2. Not divisible by 3

Case 2) When the remainder is 1 The remainder value of the expression is 1-2 = -1. Not divisible by 3

Thus, the expression will not be divisible by 3 or any multiple of it. Thus, option B.

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

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10 Sep 2017, 20:43

Since could not think of any other way used brute force to solve the question. it can be done in 2 mins.

The Min no is 29 hence we can start squaring from 6 onwards. 6²=36-5=31-C is out. 7²=49-5=44 8²=64-5=59 9²=81-5=76-D is out(38*2=76) 10²=100-5=95 11²=121-5=116-A is out (units digit is 6 try 29*4=116) 12²=144-5=139 13²=169-5=164-E is out(41*4=164) Answer is B-30

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

We need to find the answer choice that does not divide 5 less than a perfect square. Let’s analyze each answer choice:

A) 29

Since 11^2 - 5 = 116, which is divisible by 29, answer A is not correct.

B) 30

It doesn’t seem that we can find an integer q such that q^2 - 5 is divisible by 30. However, let’s make sure we can find an integer q such that q^2 - 5 is divisible by 31, 38, and 41.

C) 31

Since 6^2 - 5 = 31, which is divisible by 31, answer C is not correct.

D) 38

Since 9^2 - 5 = 76, which is divisible by 38, answer D is not correct.

E) 41

Since 13^2 - 5 = 164, which is divisible by 41, answer E is not correct.

Answer: B
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