Is (x + y)/2 an integer?

OFFICIAL SOLUTION:

Is \(\frac{x + y}{2}\) an integer?

First of all, notice that we are NOT told that \(x\) and \(y\) are integers.

(1) \(|x| = \sqrt {y^2}\)

\(|x| = |y|\). So, \(x=-y\) or \(x=y\). If \(x=-y\), then \(\frac{x + y}{2}=\frac{-y + y}{2}=0=integer\) but if \(x=y\), then \(\frac{x + y}{2}=\frac{x + x}{2}=x\), which will be an integer if \(x\) is an integer but won't be an integer if \(x\) is not an integer. Not sufficient.

(2) \(xy = 9\). If \(x=y=3\), then \(\frac{x + y}{2}=\frac{3 + 3}{2}=3=integer\) but if \(x=\frac{1}{3}\) and \(y=27\), then \(\frac{x + y}{2} \neq integer\)

(1)+(2) If from (1) \(x=-y\), then from (2) we'll have \(-x^2=9\), which is the same as \(x^2=-9\) but this is not possible (the square of a number cannot be negative). Thus, it must be true that \(x=y\), so \(x^2=9\), which gives \(x=y=3\) or \(x=y=-3\). In any case \(\frac{x + y}{2}=integer\). Sufficient.


Answer: C.