It is currently 21 Oct 2017, 03:39

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 16 May 2004
Posts: 118

Kudos [?]: 13 [0], given: 0

Location: Thailand
Are both of the integers X and Y divisible by 3? (1) X+Y and [#permalink]

Show Tags

22 May 2004, 07:31
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

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!!

Kudos [?]: 13 [0], given: 0

Manager
Joined: 10 Mar 2004
Posts: 64

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

Location: Dallas,TX

Show Tags

22 May 2004, 07:53
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

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

Manager
Joined: 16 May 2004
Posts: 118

Kudos [?]: 13 [0], given: 0

Location: Thailand

Show Tags

22 May 2004, 08:17
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!!

Kudos [?]: 13 [0], given: 0

Manager
Joined: 10 Mar 2004
Posts: 64

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

Location: Dallas,TX

Show Tags

22 May 2004, 08:26
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)

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

Manager
Joined: 16 May 2004
Posts: 118

Kudos [?]: 13 [0], given: 0

Location: Thailand

Show Tags

22 May 2004, 08:31
Maybe, we can wait other idea....Anyone please
_________________

Exceed your goals and then Proceed to Succeed!!

Kudos [?]: 13 [0], given: 0

Senior Manager
Joined: 02 Mar 2004
Posts: 327

Kudos [?]: 2 [0], given: 0

Location: There

Show Tags

22 May 2004, 11:32
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.

Kudos [?]: 2 [0], given: 0

SVP
Joined: 30 Oct 2003
Posts: 1788

Kudos [?]: 112 [0], given: 0

Location: NewJersey USA

Show Tags

23 May 2004, 08:10
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

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

Kudos [?]: 112 [0], given: 0

Manager
Joined: 16 May 2004
Posts: 118

Kudos [?]: 13 [0], given: 0

Location: Thailand

Show Tags

23 May 2004, 19:24
Thanks for anandnk's explanation.
_________________

Exceed your goals and then Proceed to Succeed!!

Kudos [?]: 13 [0], given: 0

Manager
Joined: 16 May 2004
Posts: 118

Kudos [?]: 13 [0], given: 0

Location: Thailand

Show Tags

23 May 2004, 19:27
Anyway, if n=3 and m= 3, is your explanation still applicable? Quite headache
_________________

Exceed your goals and then Proceed to Succeed!!

Kudos [?]: 13 [0], given: 0

Senior Manager
Joined: 07 Oct 2003
Posts: 349

Kudos [?]: 21 [0], given: 0

Location: Manhattan

Show Tags

23 May 2004, 19:28
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

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?

Kudos [?]: 21 [0], given: 0

Senior Manager
Joined: 02 Mar 2004
Posts: 327

Kudos [?]: 2 [0], given: 0

Location: There

Show Tags

23 May 2004, 23:47
Of course, yes.

0 = 3*0

or 0 is a multiple of any integer!

Kudos [?]: 2 [0], given: 0

23 May 2004, 23:47
Display posts from previous: Sort by