Q: Is y!/x ! an integer? (1) (x + y)(x y) = 5! + 1 (2) x + y : DS Archive
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

It is currently 08 Dec 2016, 01:53
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.

Close

Request Expert Reply

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Q: Is y!/x ! an integer? (1) (x + y)(x y) = 5! + 1 (2) x + y

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1493
Followers: 7

Kudos [?]: 96 [0], given: 0

Q: Is y!/x ! an integer? (1) (x + y)(x y) = 5! + 1 (2) x + y [#permalink]

Show Tags

New post 22 Jun 2006, 20:30
00:00
A
B
C
D
E

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.

Q: Is y!/x ! an integer?

(1) (x + y)(x – y) = 5! + 1
(2) x + y = 121
Manager
Manager
avatar
Joined: 01 Jun 2006
Posts: 140
Followers: 1

Kudos [?]: 5 [0], given: 0

 [#permalink]

Show Tags

New post 22 Jun 2006, 20:51
A it is
When we talk about X! and Y!, i think we should have x,y positive intergers.
(1) Because if (x+y)*(x-y)=5!+1 so x>y>=0
And y!/x! should not be an interger
(2) is insufficient
Intern
Intern
User avatar
Joined: 04 May 2006
Posts: 49
Followers: 0

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

 [#permalink]

Show Tags

New post 22 Jun 2006, 23:00
(1) (x + y)(x – y) = 5! + 1
(2) x + y = 121
I think it's A.

(1) (x + y)(x – y) = 5! + 1
(x + y)(x – y) = 121
i.e. (x + y)(x – y) = 121 * 1
or (x + y)(x – y) = 11 * 11

Now sum & difference of two numbers can't be same.
So, x+y = 121, x-y = 1,
Solving we get y = 60, x = 61. SO y!/x ! is not integer.
Hence 1 is SUFF.

(2) x + y = 121
Clearly not SUFF.

So A.
_________________

If A equals success, then the formula is: A = X + Y + Z, X is work. Y is play. Z is keep your mouth shut.
Albert Einstein

Director
Director
User avatar
Joined: 09 Oct 2005
Posts: 720
Location: Madrid
Followers: 3

Kudos [?]: 23 [0], given: 0

 [#permalink]

Show Tags

New post 23 Jun 2006, 00:11
mendiratta_1812 wrote:
(1) (x + y)(x ? y) = 5! + 1
(2) x + y = 121
I think it's A.

(1) (x + y)(x ? y) = 5! + 1
(x + y)(x ? y) = 121
i.e. (x + y)(x ? y) = 121 * 1
or (x + y)(x ? y) = 11 * 11

Now sum & difference of two numbers can't be same.
So, x+y = 121, x-y = 1,
Solving we get y = 60, x = 61. SO y!/x ! is not integer.
Hence 1 is SUFF.

(2) x + y = 121
Clearly not SUFF.

So A.

gr8 explanation
first was going with C
now I see it is A
_________________

IE IMBA 2010

Manager
Manager
avatar
Joined: 22 May 2006
Posts: 71
Followers: 1

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

 [#permalink]

Show Tags

New post 23 Jun 2006, 01:19
the question says about factorials and it means both x and y are postive numbers.
but,its given (x+y)*(x-y)=121.
therefore,x-y>0
this implies tht x>y,
therefore,x!/y! is an integer.
VP
VP
User avatar
Joined: 15 Jun 2006
Posts: 1124
Schools: Chicago Booth
Followers: 4

Kudos [?]: 42 [0], given: 0

 [#permalink]

Show Tags

New post 23 Jun 2006, 01:29
parvathaneni wrote:
the question says about factorials and it means both x and y are postive numbers.
but,its given (x+y)*(x-y)=121.
therefore,x-y>0
this implies tht x>y,
therefore,x!/y! is an integer.


actually the question is about y!/x!, not x!/y!, but I like your logic
Manager
Manager
avatar
Joined: 23 Jan 2006
Posts: 192
Followers: 1

Kudos [?]: 25 [0], given: 0

 [#permalink]

Show Tags

New post 23 Jun 2006, 05:23
(1) can be read as x^2 - y^2 = 121
so if x and y are positive integers (form the stem), x must be > y and y!/x! cannot be an integer
A
Current Student
User avatar
Joined: 29 Jan 2005
Posts: 5238
Followers: 24

Kudos [?]: 365 [0], given: 0

 [#permalink]

Show Tags

New post 23 Jun 2006, 06:46
kook44 wrote:
(1) can be read as x^2 - y^2 = 121
so if x and y are positive integers (form the stem), x must be > y and y!/x! cannot be an integer
A


Used the same logic and got (A)

1:10
  [#permalink] 23 Jun 2006, 06:46
Display posts from previous: Sort by

Q: Is y!/x ! an integer? (1) (x + y)(x y) = 5! + 1 (2) x + y

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| 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®.