If x, y, and z are 3 positive consecutive integers such that x < y < z

(1) (xz)^2 is even
either x or z or both are even.

We are given that there are 3 consecutive integers. Hence if x is even z has to be even.

smallest possible value of x= 2, y=3, z=4

xyz/8 will have no remainder.

(2) 5y^3 is odd

It implies that y is odd. If y is odd then x and z must be even.

And as in statement 1 xyz/8 will have no remainder.

D is the answer

OFFICIAL SOLUTION

Source : Manhattan Challenge Problems

First, note that a product of three consecutive positive integers will always be divisible by 8 if the set of these integers contains 2 even terms. These two even terms will represent consecutive multiples of 2 (note that z = x + 2), and since every other multiple of 2 is also a multiple of 4, one of these two terms will always be divisible by 4. Thus, if one of the two even terms is divisible by 4 and the other even term is divisible by 2 (since it is even), the product of 3 consecutive positive terms containing 2 even numbers will always be divisible by 8. Therefore, to address the question, we need to determine whether the set contains 2 even terms. In other words, the remainder from dividing xyz by 8 will depend on whether x is even or odd.

(1) SUFFICIENT: This statement tells us that the product xz is even. Note that since z = x + 2, x and z can be only both even or both odd. Since their product is even, it must be that both x and z are even. Thus, the product xyz will be a multiple of 8 and will leave a remainder of zero when divided by 8.

(2) SUFFICIENT: If \(5(y^3)\) is odd, then y must be odd. Since y = x +1, it must be that x = y – 1. Therefore, if y is odd, x is even, and the product xyz will be a multiple of 8, leaving a remainder of zero when divided by 8.

The correct answer is D.

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