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

Question

If x, y, and z are 3 positive consecutive integers such that x < y < z, what is the remainder when the product of x, y, and z is divided by 8?

(1) (xz)^2 is even 
(2) 5y^3 is odd

bionication · Answer

Let's try to minimize the values of X,Y,Z, in two separate scenarios:
Scenario 1: Y odd; X, Y, Z must be 2, 3, 4, respectively.
Scenario 2: Y is even; X,Y, Z must be 1, 2, 3, respectively.

(1) XZ^2 is EVEN, which aligns with Scenario 1 (XZ^2 is 64, which is even). If we multiply 2*3*4, the answer HAS to be divisible by 8 with a remainder of 0. Sufficient.

(2) 5y^3 is odd, which aligns with Scenario 1 (5*3^3 is odd x odd x odd x odd, yielding an odd outcome). Again, if we multiply 2*3*4, the answer HAS to be divisible by 8 with a remainder of 0. Sufficient.

hunterdonald · Answer

Solution: X, Y and Z can be written as X, X+1, X+2. 
Question is looking to find the remainder when (X*X+1*X+2) is divided by 2^3. 
The remainder would be 0 if the numerator has atleast three 2s

From 1) (xz)^2 is even =>    is even 
case I, x odd => (odd*odd)(odd*odd). This cannot be even. Hence, X is even 
If X is even , the numerator would have atleast 3 2s. Sufficient

From 2) 5Y*5Y*5Y is odd. Following the same logic as above, Y has to be ODD => X has to be even 
Sufficient

Answer should be D

Divyadisha · Answer

(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

BillyZ · 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.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions