If a and b are positive real numbers such that ab is an integer, is a an integer? (1) a + … — GMAT

If $a$ and $b$ are positive real numbers such that $ab$ is an integer, is $a$ an integer?

(1) $a + b$ is an integer.
(2) $a - b$ 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.

Combining both statements shows that

2a

and

2b

are integers. This restricts

a

to being either a whole integer or a half-integer. Evaluating the product of two half-integers reveals that their product has a denominator of 4 and can never be an integer. Since the question stem specifies that

ab

is an integer,

a

must be an integer.

Step-by-Step Solution

Analyze the question stem and given constraints

a

and

b

are positive real numbers and

ab = N

, where

N

is an integer. We must determine whether

a

is guaranteed to be an integer.

The variables are defined as real numbers, so we cannot assume they are integers unless forced by the mathematical statements.

Evaluate Statement (1):

a + b

is an integer

Statement (1) is NOT sufficient.

Let

a = 2 + \sqrt{3}

and

b = 2 - \sqrt{3}

. Both are positive real numbers. Here

a + b = 4

(an integer) and

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

(an integer). However,

a = 2 + \sqrt{3}

is not an integer. Conversely, if

a = 3

and

b = 1

, then

a + b = 4

and

ab = 3

, and

a

is an integer. Thus, Statement (1) alone does not uniquely determine if

a

is an integer.

Evaluate Statement (2):

a - b

is an integer

Statement (2) is NOT sufficient.

Let

a = \frac{1 + \sqrt{5}}{2}

and

b = \frac{-1 + \sqrt{5}}{2}

. Both are positive real numbers. Here

a - b = 1

(an integer) and

ab = \frac{(\sqrt{5}+1)(\sqrt{5}-1)}{4} = 1

(an integer). However,

a

is not an integer. Conversely, if

a = 4

and

b = 1

, then

a - b = 3

and

ab = 4

, and

a

is an integer. Thus, Statement (2) alone is not sufficient.

Evaluate Statement (1) and Statement (2) combined

Both statements together are SUFFICIENT.

Let

a + b = k

and

a - b = m

, where

k

and

m

are integers. Adding the two equations gives

2a = k + m \implies a = \frac{k+m}{2}

. Subtracting gives

2b = k - m \implies b = \frac{k-m}{2}

. Since

k

and

m

are integers,

k+m

and

k-m

are integers with the same parity. If

k+m

is even, both

a

and

b

are integers. If

k+m

is odd,

a = \frac{p}{2}

and

b = \frac{q}{2}

for odd integers

p

and

q

. But then

ab = \frac{pq}{4}

, which cannot be an integer because the product of two odd integers is odd and thus not divisible by 4. This contradicts the stem constraint that

ab

is an integer. Therefore,

k+m

must be even, forcing

a

to be an integer.

Key Concept

Real number vs. integer constraints and parity properties in quadratic system equations

Question

Answer

Step-by-Step Solution

Key Concept

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