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# anybody has a quick approach to this one for the value of N:

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Director
Joined: 26 Feb 2006
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anybody has a quick approach to this one for the value of N: [#permalink]  22 Apr 2007, 10:33
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anybody has a quick approach to this one for the value of N:

|NтАУ|N+1||=|NтАУ2|
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

o If n+1 >= 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N - N -1| = |N-2|
<=> 1 = |N-2|
<=> N=3 or N = 1

o If n+1 =< 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N + N+1| = |N-2|
<=> |2*N+1| = |N-2|
<=> -(2*N+1) = -(N-2) as 2*N+1 < 0 and N-2 < 0
<=> N = -3

Thus, N = -3 or N=1 or N=3
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

techjanson wrote:
Fig wrote:
o If n+1 >= 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N - N -1| = |N-2|
<=> 1 = |N-2|
<=> N=3 or N = 1

o If n+1 =< 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N + N+1| = |N-2|
<=> |2*N+1| = |N-2|
<=> -(2*N+1) = -(N-2) as 2*N+1 < 0 and N-2 < 0
<=> N = -3

Thus, N = -3 or N=1 or N=3

Excellent explanation. Thanks!

Thanks to u
Current Student
Joined: 28 Dec 2004
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Location: New York City
Schools: Wharton'11 HBS'12
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FIG: you are the abs, mod King
VP
Joined: 03 Apr 2007
Posts: 1376
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Kudos [?]: 290 [0], given: 10

Fig wrote:
o If n+1 >= 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N - N -1| = |N-2|
<=> 1 = |N-2|
<=> N=3 or N = 1

o If n+1 =< 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N + N+1| = |N-2|
<=> |2*N+1| = |N-2|
<=> -(2*N+1) = -(N-2) as 2*N+1 < 0 and N-2 < 0
<=> N = -3

Thus, N = -3 or N=1 or N=3

Can you please explain the step in red color?
SVP
Joined: 01 May 2006
Posts: 1798
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Kudos [?]: 102 [0], given: 0

fresinha12 wrote:
FIG: you are the abs, mod King

Thanks ... I just hope it helps to clarify it
SVP
Joined: 01 May 2006
Posts: 1798
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Kudos [?]: 102 [0], given: 0

goalsnr wrote:
Fig wrote:
o If n+1 >= 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N - N -1| = |N-2|
<=> 1 = |N-2|
<=> N=3 or N = 1

o If n+1 =< 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N + N+1| = |N-2|
<=> |2*N+1| = |N-2|
<=> -(2*N+1) = -(N-2) as 2*N+1 < 0 and N-2 < 0
<=> N = -3

Thus, N = -3 or N=1 or N=3

Can you please explain the step in red color?

Ok

In the second case, we have n+1 =< 0. Thus, we have |n+1| = -(n+1) : an absolute must always be positive or equal to 0.
Director
Joined: 14 Aug 2007
Posts: 734
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Kudos [?]: 124 [0], given: 0

Re: PS: Mod [#permalink]  20 Apr 2009, 00:01
Himalayan wrote:
anybody has a quick approach to this one for the value of N:

|NтАУ|N+1||=|NтАУ2|

what does NтАУ and NтАУ2 mean here?
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8

Kudos [?]: 85 [0], given: 10

Re: PS: Mod [#permalink]  20 Apr 2009, 11:19
same question as above
Manager
Joined: 22 Jul 2009
Posts: 193
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Kudos [?]: 217 [0], given: 18

Re: PS: Mod [#permalink]  02 Aug 2009, 12:29
Is there a quant guru that could explain this question in more depth?

I don't understand the transition ...

"If n+1 >= 0, then
|NтАУ|N+1||=|NтАУ2|
<=> |N - N -1| = |N-2|"

Why?

a
_________________

Please kudos if my post helps.

SVP
Joined: 29 Aug 2007
Posts: 2493
Followers: 59

Kudos [?]: 577 [0], given: 19

Re: PS: Mod [#permalink]  02 Aug 2009, 21:32
Himalayan wrote:
anybody has a quick approach to this one for the value of N:
|NтАУ|N+1||=|NтАУ2|

I guess the symbol (-ve sign) has been changed into something else. The question should be as under:

Quote:
What is the value of N:
|N - |N+1||=|N - 2|

Work out from the inside modulus
The possibilities for : (1) If (N+1) >= 0 (2) If (N+1) <= 0

(1) If (N+1) >= 0
|N - N -1| = |N - 2|
|-1| = |N - 2|
1 = |N - 2|
N = 1 or 3

(2) If (N+1) <= 0
|N+N+1| = |N - 2|
|2N+1| = |N - 2|

(i) 2N +1 = N - 2
2N - N = - 3
N = -3

(ii) 2N +1 = N - 2
2N +1 = - N + 2
2N + N = 1
N = 1/3

(ii) - 2N -1 = - N + 2
-N = 3
N = -3

N = -3, 1/3, 1, & 3.
_________________
Re: PS: Mod   [#permalink] 02 Aug 2009, 21:32
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