Easy number property problem...or is it? : PS Archive
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# Easy number property problem...or is it?

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Intern
Joined: 10 Jun 2009
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Easy number property problem...or is it? [#permalink]

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14 Jun 2009, 23:38
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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
Manager
Joined: 08 Feb 2009
Posts: 146
Schools: Anderson
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Re: Easy number property problem...or is it? [#permalink]

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15 Jun 2009, 10:29
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.
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Re: Easy number property problem...or is it? [#permalink]

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15 Jun 2009, 19:00
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?
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Re: Easy number property problem...or is it? [#permalink]

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16 Jun 2009, 06:31
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.
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Re: Easy number property problem...or is it? [#permalink]

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16 Jun 2009, 07:03
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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Re: Easy number property problem...or is it?   [#permalink] 16 Jun 2009, 07:03
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