If x is a non-zero real number, is x an integer? (1) x + \frac{6}{x} is an integer. (2) x^… — GMAT

If $x$ is a non-zero real number, is $x$ an integer?

(1) $x + \frac{6}{x}$ is an integer.

(2) $x^2 + \frac{36}{x^2}$ is an integer.

A
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.Cevap

Cevap

Statements (1) and (2) TOGETHER are NOT sufficient.

The correct response establishes that neither statement individually nor both statements combined are sufficient to determine whether the real number is an integer. Testing specific real numbers demonstrates that integer values such as 2 and irrational values such as 3 plus the square root of 3 satisfy both algebraic conditions, producing affirmative and negative answers to the question.

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Analyze Statement (1):

x + \frac{6}{x} = k

for some integer

k

x = 2

, then

2 + \frac{6}{2} = 5

(an integer), and

x

IS an integer. If

x = 3 + \sqrt{3}

, then

x + \frac{6}{x} = 3 + \sqrt{3} + \frac{6(3 - \sqrt{3})}{6} = 6

(an integer), but

x

is NOT an integer.

Since Statement (1) yields both a 'Yes' and a 'No' answer, it is not sufficient.

Analyze Statement (2):

x^2 + \frac{36}{x^2} = m

for some integer

m

x = 2

, then

2^2 + \frac{36}{2^2} = 4 + 9 = 13

(an integer), so

x

IS an integer. If

x = \sqrt{2}

, then

(\sqrt{2})^2 + \frac{36}{(\sqrt{2})^2} = 2 + 18 = 20

(an integer), but

x

is NOT an integer.

Since Statement (2) yields both a 'Yes' and a 'No' answer, it is not sufficient.

Analyze Statements (1) and (2) combined.

Note that

\left(x + \frac{6}{x}\right)^2 = x^2 + 12 + \frac{36}{x^2}

. Thus

x^2 + \frac{36}{x^2} = \left(x + \frac{6}{x}\right)^2 - 12

. Whenever

x + \frac{6}{x}

is an integer

k

x^2 + \frac{36}{x^2} = k^2 - 12

is automatically an integer. Therefore, Statement (2) provides no extra constraints.

Both

x = 2

(integer) and

x = 3 + \sqrt{3}

(non-integer) satisfy both statements simultaneously, so the combined statements remain insufficient.

Anahtar Kavram

Evaluating algebraic expressions for non-integer real number solutions in Data Sufficiency

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