## Integer N

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### Integer N

by fangtray » Wed Apr 04, 2012 5:49 am
For how many integers n is 2^n = n^2

a. none
b. one
c. two
d. three
e. more than 3

this seemed like such an easy question. i plugged in 1 found it was not one of the integers, plugged in -1 and it also was not one of them. Plugged in 0 and that did not work. Figured it was NONE

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by neelgandham » Wed Apr 04, 2012 5:54 am
fangtray wrote:For how many integers n is 2^n = n^2
a. none
b. one
c. two
d. three
e. more than 3
2^2 = 2^2
2^4 = 4^2, So the answer should be [spoiler]C = 2[/spoiler]
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by GMATGuruNY » Wed Apr 04, 2012 6:20 am
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by gmatmath » Fri Apr 06, 2012 8:12 am
We can solve this problem by trial and error method.
We will plug in values for n from 1,2,3,.... and chk for which value of n, 2^n = n^2.

1)n=1
2 = 1 FALSE

2)n=2
2^2 = 4; n^2 = 4 TRUE

3) n =3
2^3 = 8; 3^2 = 9 FALSE

4) n=4
2^4 = 16; 4^2=16 TRUE

5)n=5
2^5=32; 5^2=25 FALSE

6) n=6
2^6=64; 6^2 =36 FALSE

From this we can say that, as n increases, 2^n >> n^2.

hence for 2 values of n, this will hold true. SO the answer is option C

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by Birottam Dutta » Fri Apr 06, 2012 10:19 am
In addition to the answers above, you may also draw a graph of the two functions 2^N and N^2 and see where they intersect. There will be two points which gives the answer to the question.