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

q^2 - 5 = (q^2 - 1) - 4
= (q + 1)(q - 1) - 4

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3
if q is a multiple of 3, q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3, so it will not be a multiple of 30
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Hint: q^2-5 (q is an integer) is never multiple of 3 (try to prove this), hence 30 is out.

Answer: B.
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I did this by POE and the crude way, not sure whether we have an elegant way of solving it

IMO B,

For 29, we have option q\(= 11 Q^2=121 and 121-5 =116 = 29*4\)

for \(31 q= 6 q^2=36\)
for \(38 q= 9 q^2=81\)
for \(41 q= 13 q^2 =169\)­

q^2 - 5 = 30k + r

q^2 =30k +5 + r = 5(6k+1) + r

now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

Hence B

There is a systematic way to prove this.


That is not true, Eg. 25,55,85,.... are all multiples of 5 of the form 6k+1

Proof that \(q^2-5\) cannot be a multiple of 30

q^2 = 30k + 5 = 6*(5k) + 5
So q^2 leaves remainder 5 when divided by 6.

Consider the various cases for the remainder that q leaves when divided by 6 :
q=6k ... q^2 will leave remainder 0 when divided by 6
q=6k+1 ... q^2 will leave remainder 1 when divided by 6
q=6k+2 ... q^2 will leave remainder 4 when divided by 6
q=6k+3 ... q^2 will leave remainder 3 when divided by 6
q=6k+4 ... q^2 will leave remainder 2 when divided by 6
q=6k+5 ... q^2 will leave remainder 1 when divided by 6

So it is not possible to have a perfect square that leaves remainder 5 when divided by 6.

So 30k+5 is not a perfect square for any choice of k.
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For some integer Q, (Q^2)-5 is divisible by all of the following except
1. 29
2. 30
3. 31
4. 38
5. 41

One way is there -> POE

31+5 = 36 = 6^2 -> out

41*4 -> 164 -. 164+5 -> 13^2 -> out

38*2 = 76 -> 76+5 = 81 = 9^2 -> out

29*4 = 116 -> 116 + 5 = 121 = 11^2 -> out

Multiple all the values with 1,2,3,4 -> you will get the answer. It can be solved in 2 minutes using POE.

Only approach I could figure out to solve it quickly ( without going into details of proving wheter 30...or something )

29*1 = 29 +5 = 34
29*2= 58+5= 63
29*3 = 87+5 = 92
29*4= 116+5 = 121 ( 11^2)

41*1 = 41 +5 = 46
41*2= 82+5= 87
41*3 = 123+5 = 128
41*4= 164+5 = 169 ( 13^2)

38*1 = 38 +5 = 43
38*2= 76+5= 81 (9^2)

31*1 = 31+5 = 36 (6^2)

Left with only one choice and that is the answer.

Be strong with multiplication tables and also with the square of numbers. Then you can approach this question faster using plug and play.

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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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.

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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Bunuel

A systematic way to solve such questions is needed. Please guide.

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