Find an integer R. (1) (|R|)!=1 (2) |R|=R : DS Archive
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# Find an integer R. (1) (|R|)!=1 (2) |R|=R

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Director
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Find an integer R. (1) (|R|)!=1 (2) |R|=R [#permalink]

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13 May 2007, 20:39
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Find an integer R.

(1) (|R|)!=1
(2) |R|=R!
Manager
Joined: 02 May 2007
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13 May 2007, 21:54
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
SVP
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13 May 2007, 22:38
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
SVP
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13 May 2007, 22:40
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
Director
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13 May 2007, 22:47
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.
Director
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13 May 2007, 22:51
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.
SVP
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13 May 2007, 22:53
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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15 May 2007, 12:35
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
Director
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19 May 2007, 21:29
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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19 May 2007, 22:01
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.
19 May 2007, 22:01
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