If x and y are integers such that -8 \le x \le 4 and -2 \le y \le 5, x y^2 > 0, and x^3 y … — GMAT

If $x$ and $y$ are integers such that $-8 \le x \le 4$ and $-2 \le y \le 5$ , $x y^2 > 0$ , and $x^3 y < 0$ , how many distinct integer values are possible for the product $x y$ ?

Answer: 6

Answer

The total number of distinct integer values for the product

x y

is 6.

Analyzing the given inequalities reveals sign constraints on both variables:

x y^2 > 0

forces

x > 0

(since

y^2 > 0

for non-zero

y

), limiting

x

to integer values

1, 2, 3, 4

. Then

x^3 y < 0

forces

y < 0

(since

x^3 > 0

), limiting

y

to integer values

-2, -1

. Multiplying each value of

x

by each value of

y

gives the products

-1, -2, -3, -4, -6, -8

, which constitutes exactly 6 distinct values.

Step-by-Step Solution

Determine the sign and allowable integer values of

x

x \in \{1, 2, 3, 4\}

Since

y^2

is strictly positive for any non-zero real number

y

x y^2 > 0

implies

x > 0

and

y \neq 0

Determine the sign and allowable integer values of

y

y \in \{-2, -1\}

Because

x > 0

x^3

is also positive. For

x^3 y < 0

y

must be negative.

Calculate products for all valid pairs of

(x, y)

and remove duplicates

The distinct product values are

-1, -2, -3, -4, -6, -8

, giving 6 values in total.

Evaluating

x y

for

x \in \{1, 2, 3, 4\}

and

y \in \{-2, -1\}

yields

-1, -2, -3, -4

when

y = -1

and

-2, -4, -6, -8

when

y = -2

Key Concept

Even powers of non-zero numbers are always positive, while odd powers preserve the sign of the base number.

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