Find all School-related info fast with the new School-Specific MBA Forum

It is currently 03 Aug 2015, 20:05
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Are both of the integers X and Y divisible by 3? (1) X+Y and

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Manager
Manager
User avatar
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2

Kudos [?]: 6 [0], given: 0

Are both of the integers X and Y divisible by 3? (1) X+Y and [#permalink] New post 22 May 2004, 06:31
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Are both of the integers X and Y divisible by 3?

(1) X+Y and X-Y are divisible by 3
(2) X^2 and Y^2 are divisible by 9
_________________

Exceed your goals and then Proceed to Succeed!!

Manager
Manager
avatar
Joined: 10 Mar 2004
Posts: 64
Location: Dallas,TX
Followers: 1

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 22 May 2004, 06:53
Can be solved by each individually so ....(d)

Please note that it is already given that X and Y are intergers. If it was not specifically given then (2) may not hold good.

cheers
Manager
Manager
User avatar
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 22 May 2004, 07:17
Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2
_________________

Exceed your goals and then Proceed to Succeed!!

Manager
Manager
avatar
Joined: 10 Mar 2004
Posts: 64
Location: Dallas,TX
Followers: 1

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 22 May 2004, 07:26
becoolja wrote:
Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2


SORRYYYYYYYYYYYYYYYYYYY

was wrong above.

Cannot be solved by (1)

I hope it can be solved by (2) so (b)
Manager
Manager
User avatar
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 22 May 2004, 07:31
Maybe, we can wait other idea....Anyone please :-)
_________________

Exceed your goals and then Proceed to Succeed!!

Senior Manager
Senior Manager
avatar
Joined: 02 Mar 2004
Posts: 328
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

Re: DS_10_15 [#permalink] New post 22 May 2004, 10:32
1. x+y = 0 (3) x-y = 0 (mod 3)
or x = y = 0 (3) sufficient

2. x^2 = 0 (9), x = 0 (3)
sufficient.

D it is.
SVP
SVP
User avatar
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

Kudos [?]: 47 [0], given: 0

 [#permalink] New post 23 May 2004, 07:10
becoolja wrote:
Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2


Just extend your analysis further...

Since x and y are integers then (m+n)/2 and (m-n)/2 have to either integers
or k/3.
For no integers pairs (m,n) (m+n)/2 or (m-n)/2 yields k/3 Hence
(m+n)/2 and (m-n)/2 have to be integers. If they are integers then both x and y are divisible by 3
Manager
Manager
User avatar
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 23 May 2004, 18:24
Thanks for anandnk's explanation. :-)
_________________

Exceed your goals and then Proceed to Succeed!!

Manager
Manager
User avatar
Joined: 16 May 2004
Posts: 118
Location: Thailand
Followers: 2

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 23 May 2004, 18:27
Anyway, if n=3 and m= 3, is your explanation still applicable? Quite headache :-(
_________________

Exceed your goals and then Proceed to Succeed!!

Senior Manager
Senior Manager
User avatar
Joined: 07 Oct 2003
Posts: 357
Location: Manhattan
Followers: 2

Kudos [?]: 10 [0], given: 0

 [#permalink] New post 23 May 2004, 18:28
anandnk wrote:
becoolja wrote:
Can you explain (1) ? I am trying to solve the equation but I am not so sure it's the right method.

If x+y is divisible by 3 => x+y = 3n , when n = integer -(a)
If x-y is divisible by 3 => x-y = 3m, when m = integer -(b)

From (a) and (b) => x = 3(m+n)/2 and y = 3(n-m)/2


Just extend your analysis further...

Since x and y are integers then (m+n)/2 and (m-n)/2 have to either integers
or k/3.
For no integers pairs (m,n) (m+n)/2 or (m-n)/2 yields k/3 Hence
(m+n)/2 and (m-n)/2 have to be integers. If they are integers then both x and y are divisible by 3


Got a technical question here,
is 0 divisible by 3?
Senior Manager
Senior Manager
avatar
Joined: 02 Mar 2004
Posts: 328
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 23 May 2004, 22:47
Of course, yes.

0 = 3*0

or 0 is a multiple of any integer!
  [#permalink] 23 May 2004, 22:47
Display posts from previous: Sort by

Are both of the integers X and Y divisible by 3? (1) X+Y and

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.