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

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Director
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anybody has a quick approach to this one for the value of N: [#permalink]

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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
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22 Apr 2007, 11:53
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
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22 Apr 2007, 12:32
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
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23 Apr 2007, 12:06
FIG: you are the abs, mod King
VP
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23 Apr 2007, 12:19
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?
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23 Apr 2007, 12:25
fresinha12 wrote:
FIG: you are the abs, mod King

Thanks ... I just hope it helps to clarify it
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23 Apr 2007, 12:28
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.
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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?
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20 Apr 2009, 11:19
same question as above
Manager
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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
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Please kudos if my post helps.

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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.
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Re: PS: Mod   [#permalink] 02 Aug 2009, 21:32
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