As per your solution if X=1 then n will be 2 which is an integer.
So answer is X=1

Can you explain in detail please how to get from n= Square root(3+x) to x = 1 ????

jrakhe wrote:As per your solution if X=1 then n will be 2 which is an integer.
So answer is X=1

Can Anyone help me for further explanation in greater detail please.

phoenix,
your working is correct, got to test it with the available options.
Say, Sqr root of (3+X)

first, substitute 0 for X which would be SR of 3 - not an integer.
second value, x=2/65 results in 3 + 0.0(value) - again not an integer
if x=1, then SR of 3+1=4 is 2 - an integer.
Hence the answer is 1.

phoenix9801 wrote:Can you explain in detail please how to get from n= Square root(3+x) to x = 1 ????

From the question stem, we know that n is an integer. Accordingly, we know that sqrt(3+x) is an integer. What numbers have roots that are integers? Perfect squares.

Consequently, (3+x) must be a perfect square, i.e. 1, 4, 9, 16, 25, 36, ...

Now we just need to plug in each answer to see which ones give us perfect squares (on the GMAT you'll never see "indicate all such values" questions - is this a GRE question? Please always indicate the source!).

A) 3+0 = 3... not a perfect square.
B) 3+ (2/63) = fraction... not a perfect square.
C) 3 + 1 = 4... ding ding ding!
D) 3 + 65/63 = fraction... not a perfect square.
E) 3 + 4 = 7... not a perfect square.

Only (C) gives us a perfect square: choose C!