Interger DS

I don't think that's correct... (2) can be solved all by itself. We see that n = 3 works. Now, if you go to n = -1, -2, -3, ... the lhs gets smaller, and for n = 4, 5, 6, the lhs gets a lot bigger. So the only solution to (2) is n = 3.

So (2) is sufficient but (1) alone isn't... particularly (1)'s solution can either be -5 or 3.

Answer should be 'B'

(3 + 2) ^ 3 = 5^3

therefore n = 3

Yes I also got B

But the OA mentioned is C

I hope its wrong....

Well, are you sure it's not something like POSITIVE integers?

i think IMO is D.

from statment 1 we get,

n(n+2) = 15
n^2 + 2n - 15 =0
(n-3) (n+5)

So when we put n = 3 we get
15=15 suff.

from statment 2 we get,

(n+2)^N = 125

(n+2)^N= 5^3

so if we take n = 3

we get 5^3=5^3.

Hence D.

Please tell if i am missing something.

abhinav85 wrote:i think IMO is D.

from statment 1 we get,

n(n+2) = 15
n^2 + 2n - 15 =0
(n-3) (n+5)

So when we put n = 3 we get
15=15 suff.

from statment 2 we get,

(n+2)^N = 125

(n+2)^N= 5^3

so if we take n = 3

we get 5^3=5^3.

Hence D.

Please tell if i am missing something.

Hello, abhinav. You're missing a slight detail in the first part; n = 3, -5, and not only 3.