Author 
Message 
SVP
Joined: 03 Feb 2003
Posts: 1604

Is ! even? (1) E=Q (2) Q=2 [#permalink]
Show Tags
18 Dec 2004, 18:09
This topic is locked. If you want to discuss this question please repost it in the respective forum.
Is [E!+Q!]! even?
(1) E=Q
(2) Q=2



CIO
Joined: 09 Mar 2003
Posts: 463

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

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

the only odd factorial is 0!=1!=1, all the rest are even.
Still A?



Director
Joined: 07 Nov 2004
Posts: 683

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

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

Re: more fac [#permalink]
Show Tags
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

Yeah... I see.
I have fcked up myself. Sorry about it.
I am growing old... gettin' stupid
remember me in my golden days?



VP
Joined: 25 Nov 2004
Posts: 1483

Re: more fac [#permalink]
Show Tags
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

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

Note the factorial sign outside of the bracket rxs.



VP
Joined: 25 Nov 2004
Posts: 1483

Re: more fac [#permalink]
Show Tags
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

stolyar wrote: Yeah... I see. I have fcked 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

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

Re: more fac [#permalink]
Show Tags
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

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

nice problem, D as well
_________________
Praveen



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

It should be (d) as any factorial other than 0 and 1 is even.



Director
Joined: 07 Jun 2004
Posts: 612
Location: PA

i did not notice the outer most factorial so it should be D



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

The outermost factorial makes all the difference.







Go to page
1 2
Next
[ 23 posts ]



