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integers

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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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Source: — Problem Solving |

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.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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by arora007 » Fri Jan 28, 2011 6:29 am
thanks for posting this problem...
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