GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 15 Aug 2018, 08:25

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Senior Manager
Senior Manager
User avatar
B
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 309
For some integer q, q^2 - 5 is divisible by all of the follo  [#permalink]

Show Tags

New post 18 May 2010, 08:14
4
33
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

52% (01:51) correct 48% (02:05) wrong based on 565 sessions

HideShow timer Statistics

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
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47918
Re: q^2 - 5  [#permalink]

Show Tags

New post 18 May 2010, 09:06
5
8
Most Helpful Community Reply
Senior Manager
Senior Manager
User avatar
B
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 309
Re: q^2 - 5  [#permalink]

Show Tags

New post 08 Jan 2011, 10:05
7
3
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

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Download the Ultimate SC Flashcards

General Discussion
Senior Manager
Senior Manager
avatar
Joined: 25 Jun 2009
Posts: 285
Re: q^2 - 5  [#permalink]

Show Tags

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 \(41 q= 13 q^2 =169\)
Senior Manager
Senior Manager
User avatar
B
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 309
Re: q^2 - 5  [#permalink]

Show Tags

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 ?
Senior Manager
Senior Manager
avatar
Joined: 25 Jun 2009
Posts: 285
Re: q^2 - 5  [#permalink]

Show Tags

New post 18 May 2010, 09:04
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,
CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2653
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Reviews Badge
Re: q^2 - 5  [#permalink]

Show Tags

New post 18 May 2010, 11:18
5
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
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Manager
avatar
Joined: 17 Mar 2010
Posts: 155
Re: q^2 - 5  [#permalink]

Show Tags

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?
Manager
Manager
avatar
Joined: 20 Apr 2010
Posts: 137
Location: I N D I A
Re: q^2 - 5  [#permalink]

Show Tags

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

POE is Process of Elimination...
Manager
Manager
avatar
Joined: 05 Mar 2010
Posts: 182
Re: q^2 - 5  [#permalink]

Show Tags

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
_________________

Success is my Destiny

Manager
Manager
avatar
Joined: 14 Apr 2010
Posts: 189
Re: q^2 - 5  [#permalink]

Show Tags

New post 08 Jun 2010, 07:43
took 1 min 35 secs for plugging in :-)
Manager
Manager
avatar
Joined: 17 Mar 2010
Posts: 155
Divisibility  [#permalink]

Show Tags

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???
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 774
Location: London
GMAT ToolKit User Reviews Badge
Re: q^2 - 5  [#permalink]

Show Tags

New post 25 Sep 2010, 10:57
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.
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

GMAT Club Premium Membership - big benefits and savings

CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2653
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Reviews Badge
Re: q^2 - 5  [#permalink]

Show Tags

New post 25 Sep 2010, 11:14
:shock: must be sleeping at that time ..thanks
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Manager
avatar
Joined: 17 Mar 2010
Posts: 155
Re: q^2 - 5  [#permalink]

Show Tags

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??
Manager
Manager
avatar
Joined: 04 Jun 2010
Posts: 108
Concentration: General Management, Technology
Schools: Chicago (Booth) - Class of 2013
GMAT 1: 670 Q47 V35
GMAT 2: 730 Q49 V41
Re: q^2 - 5  [#permalink]

Show Tags

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?
_________________

Consider Kudos if my post helped you. Thanks!
--------------------------------------------------------------------
My TOEFL Debrief: http://gmatclub.com/forum/my-toefl-experience-99884.html
My GMAT Debrief: http://gmatclub.com/forum/670-730-10-luck-20-skill-15-concentrated-power-of-will-104473.html

CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2653
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Reviews Badge
Re: q^2 - 5  [#permalink]

Show Tags

New post 26 Sep 2010, 11:44
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.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Director
Director
avatar
Joined: 23 Apr 2010
Posts: 553
Re: q^2 - 5  [#permalink]

Show Tags

New post 14 Dec 2010, 06:04
How realistic is this question on the actual GMAT?
Senior Manager
Senior Manager
User avatar
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 358
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)
Re: q^2 - 5  [#permalink]

Show Tags

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.
_________________

Go Blue!

GMAT Club Premium Membership - big benefits and savings

Director
Director
User avatar
Joined: 03 Sep 2006
Posts: 836
GMAT ToolKit User
Re: q^2 - 5  [#permalink]

Show Tags

New post 14 Dec 2010, 23:49
2
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 &nbs [#permalink] 14 Dec 2010, 23:49

Go to page    1   2    Next  [ 27 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.