If k is a non-zero real number, is k an integer? (1) \frac{12}{k} is an integer. (2) k^2 -… — GMAT

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

(1) $\frac{12}{k}$ is an integer.

(2) $k^2 - 3k$ 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.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Answer
D
EACH statement ALONE is sufficient.
E
Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Both statements together are sufficient to answer the question, but neither statement alone is sufficient.

The correct response demonstrates that neither statement individually restricts

k

exclusively to integer values, but combining them enforces that

k

is rational (from the reciprocal condition) and has a denominator of 1 (from the quadratic condition), proving conclusively that

k

must be an integer.

Step-by-Step Solution

Evaluate Statement (1) independently.

Statement (1) states that

\frac{12}{k} = m

for some non-zero integer

m

, so

k = \frac{12}{m}

. If

m = 1

k = 12

(an integer). If

m = 24

k = 0.5

(not an integer). Since

k

can be either an integer or a non-integer, Statement (1) is NOT sufficient.

Testing specific values reveals that

k

can be fractional while fulfilling the condition.

Evaluate Statement (2) independently.

Statement (2) states that

k^2 - 3k = n

for some integer

n

. If

k = 4

k^2 - 3k = 4

(an integer). However, if

k = \frac{3 + \sqrt{13}}{2}

k^2 - 3k = 1

(an integer), but

k

is irrational. Since

k

can be an integer or irrational, Statement (2) is NOT sufficient.

Quadratic expressions with integer outputs can have irrational inputs.

Evaluate Statements (1) and (2) together.

From Statement (1),

k

must be rational. Let

k = \frac{p}{q}

in lowest terms, where

p

and

q

are coprimes (

\gcd(p, q) = 1

) and

q \ge 1

. Substitute

k = \frac{p}{q}

into Statement (2):

k^2 - 3k = \frac{p^2 - 3pq}{q^2}

. For this expression to be an integer,

q^2

must divide

p(p - 3q)

. Since

\gcd(p, q) = 1

q

shares no common prime factors with

p

p - 3q

. Thus

\gcd(p^2 - 3pq, q^2) = 1

, which means

q^2

can divide

p^2 - 3pq

if and only if

q^2 = 1

, so

q = 1

. Since

q = 1

k = p

, which guarantees

k

is an integer. Thus, the combined statements yield a definitive 'Yes'.

Combining rationality from the first condition with the algebraic divisibility constraint of the second condition eliminates all non-integer possibilities.

Key Concept

Data Sufficiency logic with integer constraints and rational number representations

Estimated Time:2m 0s

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Step-by-Step Solution

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