Integers: Even and Odd - Rules of Addition and Subtraction

If $$x$$, $$y$$, and $$z$$ are consecutive integers, and $$x < y < z$$, which of the following is NOT always true?

Incorrect. [[Snippet]] In order to check if this is always true, plug in even and odd values of $$x$$. If $$x = 2$$, $$y = 3$$, and $$z= 4$$, then >$$xyz =24$$; if $$x = 1$$, $$y = 2$$, and $$z = 3$$, then >$$xyz= 6$$. Since this covers both possible cases, you can be sure that $$x\cdot y\cdot z$$ must be even. __Alternative explanation__: The product of $$n$$ consecutive integers must be divisible by $$n!$$. Thus, the product of three consecutive integers must be divisible by $$3!=6$$ and must therefore be even.

Correct. [[Snippet]] In order to check if this is always true, plug in even and odd values of $$x$$. If $$x = 2$$, $$y = 3$$, and $$z = 4$$, then >$$2\cdot 3 \lt 3\cdot 4$$ >$$6 \lt 12$$. If $$x = 1$$, $$y = 2$$, and $$z = 3$$, then >$$1\cdot 2 \lt 2\cdot 3 $$ >$$2 \lt 6$$. So far so good, but is it always true for *any* number? If both $$x$$ and $$y$$ are negative, their product will be positive. For example, If $$x = -2$$, $$y = -1$$, and $$z = 0$$, then >$$xy = 2$$ and $$yz = 0$$. Based on this, $$xy$$ is not less than $$yz$$. Hence, this answer choice is *not* always true and is thus the correct answer choice.

Incorrect. [[Snippet]] In order to check if this is always true, plug in even and odd values of $$x$$. If $$x = 2$$, $$y = 3$$, and $$z = 4$$, then >$$y^2-x\cdot z = 9 - 2\cdot4 = 1$$. If $$x = 1$$, $$y = 2$$, and $$z = 3$$, then >$$y^2-x\cdot z = 4 - 3\cdot 1 = 1$$. Since this covers both possible cases, you can be sure that $$y^2-x\cdot z$$ is odd.

Incorrect. [[Snippet]] In order to check if this is always true, plug in even and odd values of $$x$$. If $$x = 1$$, $$y = 2$$, and $$z = 3$$, then >Average $$= \displaystyle \frac{1+2+3}{3}= \frac{6}{3} = 2$$. If $$x = 2$$, $$y = 3$$, and $$z = 4$$, then >Average $$= \displaystyle \frac{2+3+4}{3 }= \frac{9}{3} = 3$$. Since this covers both possible cases, you can be sure that the average of $$x$$, $$y$$, and $$z$$ must be $$y$$. __Alternative explanation__: The average of a set of consecutive terms is equal to the median, which is the number in the middle, which is $$y$$.

Incorrect. [[Snippet]] In order to check if $$yz =$$ even is always true, plug in even and odd values of $$y$$. If $$y = 2$$ and $$z = 3$$, then >$$yz = 6$$, which is even. If $$y = 1$$ and $$z = 2$$, then >$$yz = 2$$, which is also even. Since this covers both possible cases, you can be sure that $$yz$$ must be even. __Alternative explanation__: The product of $$n$$ consecutive integers must be divisible by $$n!$$. Thus, the product of two consecutive integers must be divisible by 2 and must therefore be even.

$$y\cdot z$$ is even.

The average (arithmetic mean) of $$x$$, $$y$$, and $$z$$ is $$y$$.

$$y^2-x\cdot z$$ is odd.

$$x\cdot y \lt y\cdot z$$.

$$x\cdot y\cdot z$$ is even.