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

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SVP
Joined: 03 Feb 2003
Posts: 1603

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

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18 Dec 2004, 18:02
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

Kudos [?]: 311 [0], given: 0

Senior Manager
Joined: 19 Sep 2004
Posts: 367

Kudos [?]: 6 [0], given: 0

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

Thanks
Saurabh Malpani

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CIO
Joined: 09 Mar 2003
Posts: 460

Kudos [?]: 74 [0], given: 0

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18 Dec 2004, 18:59
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.

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Senior Manager
Joined: 19 Sep 2004
Posts: 367

Kudos [?]: 6 [0], given: 0

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18 Dec 2004, 19:42
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

Kudos [?]: 6 [0], given: 0

Director
Joined: 07 Nov 2004
Posts: 683

Kudos [?]: 165 [0], given: 0

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18 Dec 2004, 23:28
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

Kudos [?]: 165 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 311 [0], given: 0

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19 Dec 2004, 00:22
E is correct

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VP
Joined: 25 Nov 2004
Posts: 1480

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19 Dec 2004, 09:23
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.

Kudos [?]: 132 [0], given: 0

19 Dec 2004, 09:23
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# A and B are integers. Is ! even? (1) A is even (2) B is even

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