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

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New post 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?

Kudos [?]: 295 [0], given: 5

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Joined: 14 Aug 2008
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Kudos [?]: 74 [2], given: 3

Schools: Harvard, Penn, Maryland
Re: Integers [#permalink]

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New post 23 Nov 2008, 15:55
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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

Kudos [?]: 74 [2], given: 3

Retired Moderator
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Joined: 18 Jul 2008
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Kudos [?]: 295 [0], given: 5

Re: Integers [#permalink]

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New post 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

Kudos [?]: 295 [0], given: 5

Manager
Manager
avatar
Joined: 02 Nov 2008
Posts: 58

Kudos [?]: 2 [0], given: 0

Re: Integers [#permalink]

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New post 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

Kudos [?]: 2 [0], given: 0

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

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