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

Incorrect.
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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.
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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.
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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.
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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.
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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.