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

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SVP
Joined: 03 Feb 2003
Posts: 1604
Is ! even? (1) E=Q (2) Q=2 [#permalink]

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18 Dec 2004, 18:09
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

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18 Dec 2004, 18:57
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: 607

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18 Dec 2004, 22:35
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: 1604

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18 Dec 2004, 22:49
the only odd factorial is 0!=1!=1, all the rest are even.
Still A?
Director
Joined: 07 Nov 2004
Posts: 683

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18 Dec 2004, 23:14
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

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19 Dec 2004, 02:09
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: 17 Oct 2004
Posts: 16

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19 Dec 2004, 05:20
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: 1604

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16 Mar 2005, 22:01
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: 1483

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16 Mar 2005, 23:58
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

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17 Mar 2005, 04:52
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: 2233

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17 Mar 2005, 09:41
Note the factorial sign outside of the bracket rxs.
VP
Joined: 25 Nov 2004
Posts: 1483

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17 Mar 2005, 18:31
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: 2233

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17 Mar 2005, 19:13
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

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18 Mar 2005, 03:54
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

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18 Mar 2005, 06:04
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

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18 Mar 2005, 14:41
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: 721
Location: Milwaukee

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20 Mar 2005, 01:09
nice problem, D as well
_________________

Praveen

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

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20 Mar 2005, 14:19
It should be (d) as any factorial other than 0 and 1 is even.
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA

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20 Mar 2005, 15:47
i did not notice the outer most factorial so it should be D
Manager
Joined: 24 Jan 2005
Posts: 217
Location: Boston

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20 Mar 2005, 16:30
The outermost factorial makes all the difference.
20 Mar 2005, 16:30

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