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# PARANOID EQUATION

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SVP
Joined: 03 Feb 2003
Posts: 1603

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31 Aug 2003, 00:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Solve the equation: (R!)!+|R|=2R

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Manager
Joined: 14 Aug 2003
Posts: 88

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Location: barcelona

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31 Aug 2003, 02:22
R! is only defined for R>=0, so we can rewrite the equation as
R!!+R=2R or R!!=R
solving we get R=1 and R=2

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SVP
Joined: 03 Feb 2003
Posts: 1603

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31 Aug 2003, 04:26
what other people think?

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Manager
Joined: 14 Aug 2003
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Location: barcelona

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31 Aug 2003, 04:38
ups! that doesnt sound as good as "correct"

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Intern
Joined: 11 Jul 2003
Posts: 3

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Location: USA

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31 Aug 2003, 10:09
R!!+R=2R -------(1)
or
R!!-R=2R---------(2)

solving (1) gives R!!=R
and solving (2) gives R!!=3R

How can anything satisfy both these equations? Cannot think of any possible values

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Intern
Joined: 17 Aug 2003
Posts: 38

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Location: USA

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31 Aug 2003, 19:29
I get the same thing as Jav.

R=1 or R=2

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Intern
Joined: 10 Aug 2003
Posts: 3

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Location: Mumbai

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02 Sep 2003, 03:06
And hey 0!! = 0 too!!!!...So, the list of values will be 0,1,2 that will satisfy this condition.

Correct me if wrong.
Anand

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Manager
Joined: 03 Jun 2003
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02 Sep 2003, 05:48
Anand Kannan Posted: Tue Sep 02, 2003 10:06 am Post subject:

--------------------------------------------------------------------------------

And hey 0!! = 0 too!!!!...So, the list of values will be 0,1,2 that will satisfy this condition.

Correct me if wrong.
Anand

0 Cannot be a correct answer. 0! = 1, 1!= 1 + 0 = 1 not equal to 2*0 = 0

For me answers 1 and 2

Kudos [?]: 1 [0], given: 0

02 Sep 2003, 05:48
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