Is ! even? (1) E=Q (2) Q=2 : Quant Question Archive [LOCKED]
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 08 Dec 2016, 19:13

### 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

# Is ! even? (1) E=Q (2) Q=2

Author Message
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 235 [0], given: 0

Is ! even? (1) E=Q (2) Q=2 [#permalink]

### Show Tags

18 Dec 2004, 17:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is [E!+Q!]! even?

(1) E=Q
(2) Q=2
CIO
Joined: 09 Mar 2003
Posts: 463
Followers: 2

Kudos [?]: 56 [0], given: 0

### Show Tags

18 Dec 2004, 17:57
I originally thought C, but I changed my mind. I think it's A. I thought originally that we had to be sure they weren't 0 or 1, because either of those factorial make 1, and 1 is odd. But knowing that the variables are equal means that even if they are 0 or 1, they both make 1, and 1 + 1 is even. So that's enough.

In 2, knowing that one variable is 2 doesn't tell us about the other one, which could be 0 or 1. So in either case, we'd get 3, which is odd, but in every other case we'd have even plus even which is even (every number other than 0 or 1 factorial must be even since it has at a minimum 2 times something in it).

So I'm going with A.
Director
Joined: 21 Sep 2004
Posts: 610
Followers: 1

Kudos [?]: 33 [0], given: 0

### Show Tags

18 Dec 2004, 21:35
ian7777 wrote:
I originally thought C, but I changed my mind. I think it's A. I thought originally that we had to be sure they weren't 0 or 1, because either of those factorial make 1, and 1 is odd. But knowing that the variables are equal means that even if they are 0 or 1, they both make 1, and 1 + 1 is even. So that's enough.

In 2, knowing that one variable is 2 doesn't tell us about the other one, which could be 0 or 1. So in either case, we'd get 3, which is odd, but in every other case we'd have even plus even which is even (every number other than 0 or 1 factorial must be even since it has at a minimum 2 times something in it).

So I'm going with A.

nice explanation Ian
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 235 [0], given: 0

### Show Tags

18 Dec 2004, 21:49
the only odd factorial is 0!=1!=1, all the rest are even.
Still A?
Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6

Kudos [?]: 142 [0], given: 0

### Show Tags

18 Dec 2004, 22:14
I would go with D.

Infact, from the stem, it looks like it will never be an odd number. You will never get 0! or 1! from E!+Q!
Besides 1! factorial every other factorial is multiplied by atleast 2 an even number and any integer multiplied by an even number will always be even.
CIO
Joined: 09 Mar 2003
Posts: 463
Followers: 2

Kudos [?]: 56 [0], given: 0

### Show Tags

19 Dec 2004, 01:09
yeah. in fact, I think it would be even no matter what. I didn't see the exclamation point after the absolute value signs originally. I saw it without that, so there was the chance of getting 2+1=3.
Intern
Joined: 16 Oct 2004
Posts: 16
Followers: 0

Kudos [?]: 1 [0], given: 0

### Show Tags

19 Dec 2004, 04:20
stolyar wrote:
Is [E!+Q!]! even?

(1) E=Q
(2) Q=2

[E!+Q!]! = (positive integer + positive integer)! = (integer >=2)! = always even
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 235 [0], given: 0

### Show Tags

16 Mar 2005, 21:01
Yeah... I see.
I have f--cked up myself. Sorry about it.

I am growing old... gettin' stupid
remember me in my golden days?
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7

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

### Show Tags

16 Mar 2005, 22:58
ChallengeMaker wrote:
stolyar wrote:
Is [E!+Q!]! even?
(1) E=Q
(2) Q=2

[E!+Q!]! = (positive integer + positive integer)! = (integer >=2)! = always even

yes. i was wondering about the question...............
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 5

Kudos [?]: 688 [0], given: 22

### Show Tags

17 Mar 2005, 03:52
my analysis is

from statement 1

(1) E=Q then 0! + 0! = 2 Or 1! + 1! = 2 and so on so A is good

(2) Q=2 here we do not know what E is so if E = 1 0r 2 then u get 1+ 2 thats odd if E = 2 thne its even so statement 2 not suff

my answers to this is A
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 320 [0], given: 0

### Show Tags

17 Mar 2005, 08:41
Note the factorial sign outside of the bracket rxs.
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7

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

### Show Tags

17 Mar 2005, 17:31
MA wrote:
ChallengeMaker wrote:
stolyar wrote:
Is [E!+Q!]! even?
(1) E=Q
(2) Q=2

[E!+Q!]! = (positive integer + positive integer)! = (integer >=2)! = always even

yes. i was wondering about the question...............

Guys, the question doesnot need any more information to answer the question. try without statement i and statement ii.
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 320 [0], given: 0

### Show Tags

17 Mar 2005, 18:13
stolyar wrote:
Yeah... I see.
I have f--cked up myself. Sorry about it.

I am growing old... gettin' stupid
remember me in my golden days?

That is a great question stolyar! And great discussions and explantions. You should drop by more often!
Intern
Joined: 30 Jan 2005
Posts: 37
Location: INDIA
Followers: 0

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

### Show Tags

18 Mar 2005, 02:54
I will go with D.
At first i missed the outside factorial sign the moment i saw i knew all number greater than 2 will give even number. So even the factorials are O or 1 you have 2.
Manager
Joined: 28 Aug 2004
Posts: 205
Followers: 1

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

### Show Tags

18 Mar 2005, 05:04
stolyar wrote:
Is [E!+Q!]! even?

(1) E=Q
(2) Q=2

C.

unless Q & E are shown to be integers, they could otherwise be irrational.
Manager
Joined: 11 Jan 2005
Posts: 57
Location: Mexico City
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

18 Mar 2005, 13:41
unless i am crazy you couldn't have factorials without using integers right?

Based on that assumption i throw in for D too.
Director
Joined: 25 Jan 2004
Posts: 728
Location: Milwaukee
Followers: 3

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

### Show Tags

20 Mar 2005, 00:09
nice problem, D as well
_________________

Praveen

Manager
Joined: 24 Jan 2005
Posts: 217
Location: Boston
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

20 Mar 2005, 13:19
It should be (d) as any factorial other than 0 and 1 is even.
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 5

Kudos [?]: 688 [0], given: 22

### Show Tags

20 Mar 2005, 14:47
i did not notice the outer most factorial so it should be D
Manager
Joined: 24 Jan 2005
Posts: 217
Location: Boston
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

20 Mar 2005, 15:30
The outermost factorial makes all the difference.

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by