If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?
I. xy(z + 1) is even.
II. x(z + y) is even.
III. (x^2 - x)yz is even.  Pick.
A) I only B) III only C) I and II D) II and III E) I, II, and III
If x is even, then the average of even number of odd consecutive integers (y), will be even. For example, the average of {1, 3} is 2.   If x is odd, then the average of odd number of odd consecutive integers (y), will be odd. For example, the average of {1, 3, 5} is 3.   Thus:   If x = even, then y = even;   If x = odd, then y = odd.   Next, |z - 6/4| = 1/2:   z - 6/4 = 1/2 --&gt; z = 2;   -(z - 6/4) = 1/2 --&gt; z = 1.   Since, z can be even as well as odd, then it takes no part in deciding whether the options are even or odd. So, we can ignore it.   I. xy(z + 1) is even: xy can be odd if both are odd and even if both are even. So, this option is not necessarily even. Discard.   II. x(z + y) is even: if x and y are odd, and z is even, then x(z + y) = odd(odd + even) = odd but if if x and y are odd, and z is odd too, then x(z + y) = odd(odd + odd) = even. So, this option is not necessarily even. Discard.   III. (x^2 - x)yz is even: (x^2 - x)yz = x(x - 1)yz. Since either x or x - 1 is even, then the whole expression must be even irrespective of the values of x, y, and z.   Answer: III only.
<result>B</result>
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?
I. xy(z + 1) is even.
II. x(z + y) is even.
III. (x^2 - x)yz is even.