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# For how many integers n is 2^n = n^2 ? A) None B) One C) Two

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Retired Moderator
Joined: 18 Jul 2008
Posts: 920
For how many integers n is 2^n = n^2 ? A) None B) One C) Two [#permalink]

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23 Nov 2008, 13:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

For how many integers n is 2^n = n^2 ?

A) None
B) One
C) Two
D) Three
E) More than Three

I plugged in numbers to get this answer. Is there another way to confirm other than plugging in numbers?
Manager
Affiliations: Beta Gamma Sigma
Joined: 14 Aug 2008
Posts: 207
Schools: Harvard, Penn, Maryland

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23 Nov 2008, 15:55
2
KUDOS
should just be 2, and 4, so C

2^2=2^2
2^4=4^2
2*2*2*2=4*4

I think this is just a simple question with the one trick that 4 also counts, as the graph of 2^x is an exponential equation that starts low and goes high, and x^2 is a quadratic equations, so after 4, the exponential's slope rises much quicker than the quadratic, which is further proof that large numbers that are unreachable by plugging in wont be found by plugging in, like 200 something. if you put it in a calculator, there are 3 intersects, including -.76something, which is not an integer. negative integers wouldnt work, obviously, because 2^-2 is 1/4.

plugging in is the best method for this question
Retired Moderator
Joined: 18 Jul 2008
Posts: 920

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24 Nov 2008, 07:51
Great explanation , +1

dk94588 wrote:
should just be 2, and 4, so C

2^2=2^2
2^4=4^2
2*2*2*2=4*4

I think this is just a simple question with the one trick that 4 also counts, as the graph of 2^x is an exponential equation that starts low and goes high, and x^2 is a quadratic equations, so after 4, the exponential's slope rises much quicker than the quadratic, which is further proof that large numbers that are unreachable by plugging in wont be found by plugging in, like 200 something. if you put it in a calculator, there are 3 intersects, including -.76something, which is not an integer. negative integers wouldnt work, obviously, because 2^-2 is 1/4.

plugging in is the best method for this question
Manager
Joined: 02 Nov 2008
Posts: 57

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24 Nov 2008, 14:15
My two cents

2^n = N^2

Can be written as N = Sqrt(2^N)

As N is integer so it can only take 2 , 4 values
Re: Integers   [#permalink] 24 Nov 2008, 14:15
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# For how many integers n is 2^n = n^2 ? A) None B) One C) Two

Moderator: chetan2u

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