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# Easy number property problem...or is it?

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Intern
Joined: 10 Jun 2009
Posts: 29
Location: Stockholm, Sweden
Easy number property problem...or is it? [#permalink]

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15 Jun 2009, 00:38
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For how many integers N is 2^N = N^2

1) None
2) One
3) Two
4) Three
5) More than three

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Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson
Re: Easy number property problem...or is it? [#permalink]

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15 Jun 2009, 11:29
PhilosophusRex wrote:
For how many integers N is 2^N = N^2

1) None
2) One
3) Two
4) Three
5) More than three

For N = 2 & 4, the equation is satisfied.
Intern
Joined: 08 Jun 2009
Posts: 32
Re: Easy number property problem...or is it? [#permalink]

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15 Jun 2009, 20:00
goldeneagle94 wrote:

1) None
2) One
3) Two
4) Three
5) More than three

For N = 2 & 4, the equation is satisfied.

Is there any number property associated? Or is it by trial and error?
Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson
Re: Easy number property problem...or is it? [#permalink]

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16 Jun 2009, 07:31
Jozu wrote:
goldeneagle94 wrote:

1) None
2) One
3) Two
4) Three
5) More than three

For N = 2 & 4, the equation is satisfied.

Is there any number property associated? Or is it by trial and error?

If you plug-in numbers from -2 to 6, you will see a trend. After which point, there is no need to plug-in because LHS of the equation starts to turn into an enormous value.
Manager
Joined: 16 Apr 2009
Posts: 219
Schools: Ross
Re: Easy number property problem...or is it? [#permalink]

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16 Jun 2009, 08:03
Quote:
For how many integers N is 2^N = N^2

1) None
2) One
3) Two
4) Three
5) More than three

Good question.
2^2=2^2
2^4=4^2
IMO ,OA is C

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: Easy number property problem...or is it?   [#permalink] 16 Jun 2009, 08:03
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