If r is a positive real number, is r an integer? (1) r^2 + 2r is an integer. (2) r^2 - 2r … — GMAT

If $r$ is a positive real number, is $r$ an integer?

(1) $r^2 + 2r$ is an integer.
(2) $r^2 - 2r$ 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, but NEITHER statement ALONE is sufficient.

Neither statement alone is sufficient because irrational numbers like

\sqrt{2}-1

and

\sqrt{2}+1

can satisfy statements (1) and (2) respectively. However, combining both statements proves that

4r

is an integer

k

, and substituting

r = k/4

into the quadratic equation shows that

k

must be a multiple of 4, forcing

r

to be a positive integer.

Step-by-Step Solution

Evaluate Statement (1) independently.

Statement (1) is INSUFFICIENT.

r = 1

, then

r^2 + 2r = 3

(an integer), and

r

is an integer (YES). If

r = \sqrt{2} - 1

, then

r^2 + 2r = (3 - 2\sqrt{2}) + (2\sqrt{2} - 2) = 1

(an integer), but

r

is NOT an integer (NO). Since

r

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

Evaluate Statement (2) independently.

Statement (2) is INSUFFICIENT.

r = 3

, then

r^2 - 2r = 3

(an integer), and

r

is an integer (YES). If

r = \sqrt{2} + 1

, then

r^2 - 2r = (3 + 2\sqrt{2}) - (2\sqrt{2} + 2) = 1

(an integer), but

r

is NOT an integer (NO). Since

r

can be an integer or a non-integer, Statement (2) alone is not sufficient.

Evaluate Statement (1) and Statement (2) together.

Statements (1) and (2) together are SUFFICIENT.

Let

r^2 + 2r = a

and

r^2 - 2r = b

, where

a

and

b

are integers. Subtracting Statement (2) from Statement (1) gives

(r^2 + 2r) - (r^2 - 2r) = a - b \implies 4r = a - b

. Since

a

and

b

are integers,

a - b = k

for some integer

k

. Thus,

r = \frac{k}{4}

. Now substitute

r = \frac{k}{4}

back into Statement (1):

\left(\frac{k}{4}\right)^2 + 2\left(\frac{k}{4}\right) = a \implies \frac{k^2}{16} + \frac{k}{2} = a \implies \frac{k^2 + 8k}{16} = a

. This means

16

must divide

k^2 + 8k = k(k+8)

. If

k

were odd,

k(k+8)

would be odd and thus not divisible by 16. So

k

must be even. Let

k = 2m

. Then

k(k+8) = 2m(2m+8) = 4m(m+4)

. For

16

to divide

4m(m+4)

4

must divide

m(m+4)

. If

m

were odd,

m(m+4)

would be odd, which is not divisible by 4. Thus,

m

must be even. Let

m = 2p

. Then

k = 2m = 4p

, which means

k

is a multiple of 4. Therefore,

r = \frac{k}{4} = \frac{4p}{4} = p

, an integer. Thus,

r

MUST be an integer.

Key Concept

Testing integer constraints in Data Sufficiency requires evaluating whether non-integer real numbers satisfy given equations, and combining equations to determine if rational and integer bounds strictly hold.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic