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Manager
Joined: 18 Mar 2004
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01 Apr 2007, 13:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is B>=0?

1) B * |B| = B^2
2)|B| + B!=2

What is the answer and why?

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SVP
Joined: 01 May 2006
Posts: 1794

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01 Apr 2007, 13:27
(D) for me

B >= 0

From 1
B * |B| = B^2

As B^2 >= 0 by definition, we have:
B * |B| >= 0
<=> B >= 0 as |B| >= 0 by definition too.

SUFF.

From 2
|B| + B!=2 >>> By definition of the factorial, B must be a positive integer or 0.

SUFF.

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Director
Joined: 30 Nov 2006
Posts: 591

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Location: Kuwait

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01 Apr 2007, 13:31
The question asks whether B is either a positive number or zero

Statement 1: B * |B| = B^2

The only case for this is when B is either positive becuase B^2 must be positive So both B and |B| are either positive or negative OR when B is zero and zero x zero = zero. Since |B|, by definition, is always positive --> B is positive or zero.

statement 1 is sufficient

Statement 2: |B| + B!=2

|B| is by definition positive. Because |B| + B! result in a positive integer, B! can either be a negative number that is smaller than B or a positive number. By definition, B! is positive non-zero number because factorials are valid of positive numbers only and the resulting factorial can never equal zero. Therefore, YES, B is a positive number.

Statement 2 is sufficient

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Senior Manager
Joined: 20 Feb 2007
Posts: 256

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02 Apr 2007, 00:53
Quote:
Is B>=0?

1) B * |B| = B^2
2)|B| + B!=2

Taking (1):
B * |B| = B^2 = B*B

B * |B| = B*B

|B| = B*B/B = B

|B| = B Since |B| is always +ve so B is a positive integer SUFF

(2) Fig wrote:
Quote:
From 2
|B| + B!=2 >>> By definition of the factorial, B must be a positive integer or 0. SUFF.

Earlier, I was confused with stmt (2) but thanks Fig!

Last edited by Summer3 on 02 Apr 2007, 01:09, edited 1 time in total.

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Director
Joined: 13 Dec 2006
Posts: 506

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Location: Indonesia

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02 Apr 2007, 01:02
D from me

regards,

Amardeep

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02 Apr 2007, 01:02
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