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

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

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New post 18 May 2010, 08:14
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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
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Re: q^2 - 5  [#permalink]

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New post 18 May 2010, 09:06
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Re: q^2 - 5  [#permalink]

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New post 08 Jan 2011, 10:05
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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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Re: q^2 - 5  [#permalink]

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New post 18 May 2010, 08:26
[/m]
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


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 \(41 q= 13 q^2 =169\)
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Re: q^2 - 5  [#permalink]

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New post 18 May 2010, 08:57
nitishmahajan wrote:
[/m]
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


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 \(41 q= 13 q^2 =169\)


hmmm ....im wondering how long you took to solve this ...?...do you think we can generalize this method and use it ?
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Re: q^2 - 5  [#permalink]

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New post 18 May 2010, 09:04
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dimitri92 wrote:
nitishmahajan wrote:
[/m]
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


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 \(41 q= 13 q^2 =169\)


hmmm ....im wondering how long you took to solve this ...?...do you think we can generalize this method and use it ?



It took me around 3 minutes or so but I dont think this is the best way of solving this,
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Re: q^2 - 5  [#permalink]

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New post 18 May 2010, 11:18
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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
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Re: q^2 - 5  [#permalink]

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New post 19 May 2010, 03:15
Can somebody post the answer in the systematic explained way?? For the sake of new and weak users?
I mean how do you start to think when you see this problem. The systematic approach can be developed for this kind of problems??
And by the way what is POE?
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Re: q^2 - 5  [#permalink]

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New post 21 May 2010, 02:11
@amitjash...

POE is Process of Elimination...
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Re: q^2 - 5  [#permalink]

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New post 22 May 2010, 03:30
IMO B

i just plugged in the values of q from 6 onwards and it took me less than 2 minutes
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Re: q^2 - 5  [#permalink]

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New post 08 Jun 2010, 07:43
took 1 min 35 secs for plugging in :-)
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Divisibility  [#permalink]

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New post 25 Sep 2010, 10:03
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

Can some one explain how to solve this???
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Re: q^2 - 5  [#permalink]

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New post 25 Sep 2010, 10:57
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There is a systematic way to prove this.

gurpreetsingh wrote:
q^2 - 5 = 30k + r
now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

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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Re: q^2 - 5  [#permalink]

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New post 25 Sep 2010, 11:14
:shock: must be sleeping at that time ..thanks
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Re: q^2 - 5  [#permalink]

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New post 26 Sep 2010, 10:00
I am sorry but still i am not sure if i face this problem in exam i will be able to crack this.. How you will target at 30?? I mean you have all other options around...??? Is there any way to crack this type of problems??
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Re: q^2 - 5  [#permalink]

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New post 26 Sep 2010, 11:36
shrouded1 wrote:
There is a systematic way to prove this.

gurpreetsingh wrote:
q^2 - 5 = 30k + r
now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

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.


This is a good explanation but how do you apply this on the other answers quickly?
30 is easily divided to 5 and 6. but what about 41 or 29? you have to think about a number which these numbers are multiplied by and than add 5 which is a perfect square. Not trivial at all I think.

Any shortcut suggestions?
What's the source of the question?
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Re: q^2 - 5  [#permalink]

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New post 26 Sep 2010, 11:44
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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.
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Re: q^2 - 5  [#permalink]

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New post 14 Dec 2010, 06:04
How realistic is this question on the actual GMAT?
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Re: q^2 - 5  [#permalink]

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New post 14 Dec 2010, 08:19
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.


This is a great tip.

I was plugging in values too, but this could save me a good 2 min.
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Re: q^2 - 5  [#permalink]

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New post 14 Dec 2010, 23:49
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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.
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Re: q^2 - 5   [#permalink] 14 Dec 2010, 23:49

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