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integers

by Ramit88 » Tue Jan 25, 2011 10:17 pm
For how many integers n 2^n = n^2

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

[spoiler]
ANS C[/spoiler]

acc to me its B i.e one

when n =2,
2^n = n^2
2^2 = 2 ^2

it can not be zero

2^0 = 0^2
1 = not defined

pls point out my mistake
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by Anurag@Gurome » Tue Jan 25, 2011 10:39 pm
Ramit88 wrote:For how many integers n 2^n = n^2

a none
b one
c two
d three
e more than 3
There two possible values of n satisfying the relation 2^n = n^2
  • 1. n = 2 --> 2^2 = 4 = 2^2
    2. n = 4 --> 2^4 = 16 = 4^2
The correct answer is C.
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by prachich1987 » Fri Jan 28, 2011 3:41 am
Anurag@Gurome wrote:
Ramit88 wrote:For how many integers n 2^n = n^2

a none
b one
c two
d three
e more than 3
There two possible values of n satisfying the relation 2^n = n^2
  • 1. n = 2 --> 2^2 = 4 = 2^2
    2. n = 4 --> 2^4 = 16 = 4^2
The correct answer is C.
Hi Anurag,

Is there any algebraic approach to this.

Thanks!
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by Anurag@Gurome » Fri Jan 28, 2011 3:46 am
prachich1987 wrote:Is there any algebraic approach to this.
Yes, there is.
But unless you have a strong graphical sense and good understanding of logarithm, I discourage you to go for that approach. For GMAT purpose, just remember that there is only two such pair and also remember the values.
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by arora007 » Fri Jan 28, 2011 6:29 am
thanks for posting this problem...
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