A and B are integers. Is ! even? (1) A is even (2) B is even : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 10:12

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

A and B are integers. Is ! even? (1) A is even (2) B is even

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

Kudos [?]: 245 [0], given: 0

A and B are integers. Is ! even? (1) A is even (2) B is even [#permalink]

Show Tags

18 Dec 2004, 17:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A and B are integers. Is [A+B]! even?

(1) A is even

(2) B is even
Senior Manager
Joined: 19 Sep 2004
Posts: 369
Followers: 1

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

Show Tags

18 Dec 2004, 17:11
I will opt for D.

Thanks
Saurabh Malpani
CIO
Joined: 09 Mar 2003
Posts: 463
Followers: 2

Kudos [?]: 58 [0], given: 0

Show Tags

18 Dec 2004, 17:59
stolyar wrote:
A and B are integers. Is [A+B]! even?

(1) A is even

(2) B is even

I've seen this [ xxx ] before on this board, but I have no idea what it means. I've seen people call it "mod xxx". Is that right? What is it? Is it supposed to be absolute value?

Sorry for not knowing - I just am unsure.

If it's absolute value, then I guess I think we just need to make sure that the numbers don't come out to 1 or 0. And even together we can plug in something like 2 and -2, which are both even, but together make 0. And 0! is 1, which is odd.

So I'm going with E.
Senior Manager
Joined: 19 Sep 2004
Posts: 369
Followers: 1

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

Show Tags

18 Dec 2004, 18:42
Hey1 Ian yes You are CORRECT. I forgot abour -ve numbers how stupid of me.

As soon as I saw factorial I was like +ve numbers. You are absolutely correct.

Thanks
Saurabh Malpani
Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6

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

Show Tags

18 Dec 2004, 22:28
I would pick E.

s1: A is even
We dont know anything about B. Insufficient

s1: B is Even
We dont know anything about B. Insufficient

s1+s2: Since it does not also say that A & B are distinct, if A=B=0
then [A+B]! = 0! = 1 odd
or if A=2, B=0
then [A+B]! = 2! = 2 even
Insufficient
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 245 [0], given: 0

Show Tags

18 Dec 2004, 23:22
E is correct
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7

Kudos [?]: 98 [0], given: 0

Show Tags

19 Dec 2004, 08:23
gayathri wrote:
I would pick E.

s1: A is even
We dont know anything about B. Insufficient

s1: B is Even
We dont know anything about B. Insufficient

s1+s2: Since it does not also say that A & B are distinct, if A=B=0
then [A+B]! = 0! = 1 odd
or if A=2, B=0
then [A+B]! = 2! = 2 even
Insufficient

Agree with E. if A nad B are negative even integers, then [A+B]! can not even be defined.
19 Dec 2004, 08:23
Display posts from previous: Sort by