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

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42 00:00

Difficulty:   95% (hard)

Question Stats: 48% (02:20) correct 52% (02:28) wrong based on 625 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58340
Re: q^2 - 5  [#permalink]

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5
9
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.

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Re: q^2 - 5  [#permalink]

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7
5
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
General Discussion
Manager  Joined: 25 Jun 2009
Posts: 234
Re: q^2 - 5  [#permalink]

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[/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 $$41 q= 13 q^2 =169$$
Senior Manager  B
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Re: q^2 - 5  [#permalink]

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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 ?
Manager  Joined: 25 Jun 2009
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Re: q^2 - 5  [#permalink]

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1
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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GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: q^2 - 5  [#permalink]

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6
1
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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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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@amitjash...

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

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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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took 1 min 35 secs for plugging in Manager  Joined: 17 Mar 2010
Posts: 130

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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

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

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1
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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GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: q^2 - 5  [#permalink]

Show Tags must be sleeping at that time ..thanks
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Re: q^2 - 5  [#permalink]

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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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GMAT 1: 670 Q47 V35 GMAT 2: 730 Q49 V41 Re: q^2 - 5  [#permalink]

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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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GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: q^2 - 5  [#permalink]

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2
3
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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How realistic is this question on the actual GMAT?
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Re: q^2 - 5  [#permalink]

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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.

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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4
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. Re: q^2 - 5   [#permalink] 14 Dec 2010, 23:49

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