If x and y are positive real numbers, is x an integer? (1) x + y is an integer and xy = 12… — GMAT

If $x$ and $y$ are positive real numbers, is $x$ an integer?

(1) $x + y$ is an integer and $xy = 12$ .
(2) $x - y$ is an integer and $x^2 + y^2 = 25$ .

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 determine that

x

must be an integer, but neither statement alone is sufficient.

Evaluating both statements together establishes that

(x+y)^2 = x^2 + y^2 + 2xy = 25 + 2(12) = 49

, meaning

x+y=7

. The system

x+y=7

and

xy=12

yields solutions

x=3, y=4

x=4, y=3

. In both outcomes,

x

is guaranteed to be an integer, yielding a definitive 'Yes' answer.

Step-by-Step Solution

Evaluate Statement (1) independently.

Statement (1) gives

x + y = k

for some integer

k

, and

xy = 12

. If

k = 7

, the quadratic equation

t^2 - 7t + 12 = 0

gives roots

t = 3

and

t = 4

, which are integers. However, if

k = 9

, the quadratic equation

t^2 - 9t + 12 = 0

gives roots

t = \frac{9 \pm \sqrt{33}}{2}

. Here

x = \frac{9 + \sqrt{33}}{2}

is a positive real number,

y = \frac{9 - \sqrt{33}}{2} > 0

x+y = 9

(integer), and

xy = 12

, but

x

is NOT an integer. Thus,

x

can be an integer or a non-integer.

Demonstrate that positive real numbers whose sum and product are integers do not necessarily have to be integers themselves.

Evaluate Statement (2) independently.

Statement (2) gives

x - y = m

for some integer

m

, and

x^2 + y^2 = 25

. If

m = 1

, we can have

x = 4

and

y = 3

, so

x

is an integer. If

m = 2

, then

(x-y)^2 = x^2 - 2xy + y^2 \implies 4 = 25 - 2xy \implies 2xy = 21

. Then

(x+y)^2 = x^2 + y^2 + 2xy = 25 + 21 = 46 \implies x+y = \sqrt{46}

. Thus

x = \frac{\sqrt{46} + 2}{2}

, which is positive and real, but NOT an integer. Thus, Statement (2) alone is not sufficient.

Test whether integer difference and fixed sum of squares guarantees integer values.

Evaluate Statements (1) and (2) together.

From Statement (1),

xy = 12

and

x + y = k

(where

k

is an integer). From Statement (2),

x^2 + y^2 = 25

. Using the algebraic identity

x^2 + y^2 = (x+y)^2 - 2xy

, we substitute the known values:

25 = k^2 - 2(12) \implies 25 = k^2 - 24 \implies k^2 = 49

. Since

x > 0

and

y > 0

x + y = k > 0

, so

k = 7

. Now we have

x + y = 7

and

xy = 12

. The possible values for

x

and

y

are the solutions to

t^2 - 7t + 12 = 0 \implies (t-3)(t-4) = 0 \implies t \in \{3, 4\}

. Therefore,

x

must be either 3 or 4. In both cases,

x

is definitively an integer.

Combine the conditions to solve for

x+y

and prove

x

must be an integer.

Key Concept

Integer constraints vs. real number assumptions in Data Sufficiency algebraic systems

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