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

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

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