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

[pleasehelp]

the answer is two. i answered 0.

which two? :)

[crackgmat]

For n = 2 and n = 4

Thanks

[Akshay]

2 values of n...

n=2 & 4...

Hope m right... :roll: :roll:

[RonPurewal]

yes, 2 and 4.
hopefully, 2 jumped out at you pretty fast; if n = 2, then the two sides of the equation are 2^2 and 2^2. i'm pretty sure that those are going to be equal.
you'll discover n = 4 through raw experimentation; there isn't any better way, unfortunately.

if you want to be confident that these are the only solutions, you have to watch the behavior of 2^n and n^2 as you get further and further away from 4. the pattern you'll observe is that 2^n begins to grow much, much faster than does n^2, making it clear that the two expressions won't be equal for any larger values.
the equality is definitely impossible for negative integers, because 2^(negative integer) is a fraction, while (negative integer)^2 is not. therefore, you don't have to worry about negative integers.

[be.a.true.winner]

yes, 2 and 4.

the equality is definitely impossible for negative integers, because 2^(negative integer) is a fraction, while (negative integer)^2 is not. therefore, you don't have to worry about negative integers.

Just to confirm, I think if the question did not mention "integers", there would be 3 possible solutions, right Ron?

[RonPurewal]

yes, 2 and 4.

the equality is definitely impossible for negative integers, because 2^(negative integer) is a fraction, while (negative integer)^2 is not. therefore, you don't have to worry about negative integers.

Just to confirm, I think if the question did not mention "integers", there would be 3 possible solutions, right Ron?

yeah, there would be another solution (some number between -1 and 0). but that is way, way beyond the scope of the gmat.

[sachin.w]

yes, 2 and 4.
hopefully, 2 jumped out at you pretty fast; if n = 2, then the two sides of the equation are 2^2 and 2^2. i'm pretty sure that those are going to be equal.
you'll discover n = 4 through raw experimentation; there isn't any better way, unfortunately.

if you want to be confident that these are the only solutions, you have to watch the behavior of 2^n and n^2 as you get further and further away from 4. the pattern you'll observe is that 2^n begins to grow much, much faster than does n^2, making it clear that the two expressions won't be equal for any larger values.
the equality is definitely impossible for negative integers, because 2^(negative integer) is a fraction, while (negative integer)^2 is not. therefore, you don't have to worry about negative integers.

hi ron,
I tested till 3 and stopped and thought there can only be 1 integer possible..
had i tried 4 as well, i would have got the rigth ans..

my question is how do i know when to stop testing and till when do i need to go on testing.
kindly help.

[RonPurewal]

my question is how do i know when to stop testing and till when do i need to go on testing.
kindly help.

you should keep testing until a pattern becomes clear. if there's no discernible pattern to what is happening, you should keep throwing numbers in there until there is one.

here, if the only numbers you plugged were 1, 2, and 3, there's no way you would have discovered a meaningful pattern.
if you kept plugging past 4, you would have seen rather quickly that 2^n starts growing much faster than n^2, thus guaranteeing that there would be no more solutions.

[sachin.w]

This, for sure, is an amazing piece of advice. Thanks a ton, Ron for all the help.

[jnelson0612]

This, for sure, is an amazing piece of advice. Thanks a ton, Ron for all the help.

Glad that you received the help you needed. :-)