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Director
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Find an integer R. (1) (|R|)!=1 (2) |R|=R [#permalink]
 13 May 2007, 21:39
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Find an integer R.
(1) (|R|)!=1
(2) |R|=R!
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Manager
Joined: 02 May 2007
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Himalayan wrote: Find an integer R.
(1) (|R|)!=1
(2) |R|=R!
(C) it is
(1) => R can be 1 or -1 => insuff
(2) => R can be 1 or 2 => insuff
(1) & (2) => R = 1
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SVP
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kirakira wrote: Himalayan wrote: Find an integer R.
(1) (|R|)!=1
(2) |R|=R! (C) it is (1) => R can be 1 or -1 => insuff (2) => R can be 1 or 2 => insuff (1) & (2) => R = 1 (C) as well, I agree 
To complete the stat 1 analysis, we have also R=0 as a possibility. 0! = 1
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SVP
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Himalayan wrote: kirakira wrote: Himalayan wrote: Find an integer R.
(1) (|R|)!=1
(2) |R|=R! (1) => R can be 1 or -1 => insuff why R is only 1 and -1? 1! = 1 and |1| = |-1| = 1
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Director
Joined: 12 Jun 2006
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Fig wrote: kirakira wrote: Himalayan wrote: Find an integer R.
(1) (|R|)!=1
(2) |R|=R! (C) it is (1) => R can be 1 or -1 => insuff (2) => R can be 1 or 2 => insuff (1) & (2) => R = 1 (C) as well, I agree To complete the stat 1 analysis, we have also R=0 as a possibility. 0! = 1 In (2), isn't |R| = -R or R? If so, how can -R = R!. Can we "factorialize" a negative #? Same question w/stmnt. 1. Isn't (|R|) = (R) or (-R). I'm prob missing something.
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Director
Joined: 26 Feb 2006
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ggarr wrote: Fig wrote: kirakira wrote: Himalayan wrote: Find an integer R.
(1) (|R|)!=1
(2) |R|=R! (C) it is (1) => R can be 1 or -1 => insuff (2) => R can be 1 or 2 => insuff (1) & (2) => R = 1 (C) as well, I agree To complete the stat 1 analysis, we have also R=0 as a possibility. 0! = 1 In (2), isn't |R| = -R or R? If so, how can -R = R!. Can we "factorialize" a negative #? Same question w/stmnt. 1. Isn't (|R|) = (R) or (-R). I'm prob missing something. -ve integers do not have factorials.
in 1, R can be -1, 0 and 1.
C is correct.
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SVP
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ggarr wrote: Fig wrote: kirakira wrote: Himalayan wrote: Find an integer R.
(1) (|R|)!=1
(2) |R|=R! (C) it is (1) => R can be 1 or -1 => insuff (2) => R can be 1 or 2 => insuff (1) & (2) => R = 1 (C) as well, I agree To complete the stat 1 analysis, we have also R=0 as a possibility. 0! = 1 In (2), isn't |R| = -R or R? If so, how can -R = R!. Can we "factorialize" a negative #? Same question w/stmnt. 1. Isn't (|R|) = (R) or (-R). I'm prob missing something. To me, I read this problem like this:
For 1: (|R|)! = 1
> (|1|)! = 1! = 1
> (|0|)! = 0! = 1
> (|-1|)! = 1! = 1
For 2: |R| = R!
> |1| = 1! = 1
> |2| = 2! = 2
We never have a factorization of a negative number
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Current Student
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I am going with C..
R?
|R|!=1, well if R=0, |-1|, 1, the factorial of these numbers=1
|R|=R!, well if R=1, or 2..then we meet the requirement
now R! is positive..so which implies..R has to be positive..therefore
taking both statment 1) and 2) together we know R=1
C it is
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Director
Joined: 26 Feb 2006
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Fig wrote: kirakira wrote: Himalayan wrote: Find an integer R.
(1) (|R|)!=1
(2) |R|=R! (C) it is (1) => R can be 1 or -1 => insuff (2) => R can be 1 or 2 => insuff (1) & (2) => R = 1 (C) as well, I agree To complete the stat 1 analysis, we have also R=0 as a possibility. 0! = 1 you got C on completing this one. 
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CEO
Joined: 29 Mar 2007
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Ya its C. |R|= -R or R. So stmnts 1 and 2 are insuff.
But together since they both have 1 as a value, then its true... Well thats how the Kaplan book explains these things.
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