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if n^2 is integer, it means nothing because if n^2 is negative, n is non-integer. Now if sqrt(n) is integer, n is (not only) integer(, but non-negative. So they should have asked if n is non-negative integer. But the) answer to the question is B. because 2 is sufficient and 1 is not.

GmatEnemy wrote:

Is n an integer ?

1. n^2 is an integer. 2. Sqrt (n) is an integer

I dont understand what the hell is the trap in this problem ? Its soo simple but i am shocked with the answer.

if n^2 is integer, it means nothing because if n^2 is negative, n is non-integer. Now if sqrt(n) is integer, n is (not only) integer(, but non-negative. So they should have asked if n is non-negative integer. But the) answer to the question is B. because 2 is sufficient and 1 is not.

GmatEnemy wrote:

Is n an integer ?

1. n^2 is an integer. 2. Sqrt (n) is an integer

I dont understand what the hell is the trap in this problem ? Its soo simple but i am shocked with the answer.

Tushar, how can n^2 be negative? I did not get that part. Also, negative numbers too are integers.
_________________

if n^2 is integer, it means nothing because if n^2 is negative, n is non-integer. Now if sqrt(n) is integer, n is (not only) integer(, but non-negative. So they should have asked if n is non-negative integer. But the) answer to the question is B. because 2 is sufficient and 1 is not.

GmatEnemy wrote:

Is n an integer ?

1. n^2 is an integer. 2. Sqrt (n) is an integer

I dont understand what the hell is the trap in this problem ? Its soo simple but i am shocked with the answer.

n^2 can never be negative. the lowest possible value of n^2 is 0. also an integer can be negative.

For an integer (positive or negative) n, n^2 is always positive. But 1 gives only that n^2 is integer and not non-negative integer; so if n^2 is negative, n is not integer. Thus, 1 is not enough. Next 2=>sqrt(n)=a=>a^2=n; Thus, n is non-negative i.e. 0 or positive. Thus, 2 is sufficient.

krishan wrote:

tusharvk wrote:

if n^2 is integer, it means nothing because if n^2 is negative, n is non-integer. Now if sqrt(n) is integer, n is (not only) integer(, but non-negative. So they should have asked if n is non-negative integer. But the) answer to the question is B. because 2 is sufficient and 1 is not.

GmatEnemy wrote:

Is n an integer ?

1. n^2 is an integer. 2. Sqrt (n) is an integer

I dont understand what the hell is the trap in this problem ? Its soo simple but i am shocked with the answer.

Tushar, how can n^2 be negative? I did not get that part. Also, negative numbers too are integers.

A does not say whether n is a positive or negative, hence, (A) is not sufficient.

No, all numbers on the GMAT are real numbers. You can never take square roots of negatives on the GMAT. If you see \(n^2\) on the GMAT, it is always de facto true that \(n^2 \geq 0\); it is never relevant to a GMAT problem to consider the possibility that \(n^2\) is negative.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

if N^2 is always non-negative for GMAT, does that change the answer. Your point is well taken; I do not think it will change the answer. n^2=2 is integer; but n is non-integer. sqrt(n) is integer=>n which is square of integer is integer and also n is non-negative.

IanStewart wrote:

krishan wrote:

oh !!! got it .

if n = sqrt(-2) , n^2 = -2.

A does not say whether n is a positive or negative, hence, (A) is not sufficient.

No, all numbers on the GMAT are real numbers. You can never take square roots of negatives on the GMAT. If you see \(n^2\) on the GMAT, it is always de facto true that \(n^2 \geq 0\); it is never relevant to a GMAT problem to consider the possibility that \(n^2\) is negative.

if N^2 is always non-negative for GMAT, does that change the answer. Your point is well taken; I do not think it will change the answer. n^2=2 is integer; but n is non-integer. sqrt(n) is integer=>n which is square of integer is integer and also n is non-negative.

The original answer is correct, but not because n^2 might be negative. The original answer is correct because n^2 might be equal to a positive integer which is not a perfect square.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com