japped187 wrote:
If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?
(1) wx + yz is odd
(2) wz + yx is odd
\(\frac{w}{x} + \frac{y}{z} = \frac{wz + xy}{xz}\)
Since \(\frac{w}{x}\) and \(\frac{y}{z}\) are integers, their SUM on the left side of the equation above must also be an integer.
Implication:
\(\frac{wz + xy}{xz}\) = INTEGER
Statement 1: wx + yz = odd
Case 1: w=1, x=1, y=2 and z=2, with the result that wx + yz = 1*1 + 2*2 = 5
In this case, \(\frac{w}{x} +\frac{y}{z} = \frac{1}{1} + \frac{2}{2} = 2\), so the answer to the question stem is NO.
Case 2: w=1, x=1, y=2, and z=1, with the result that wx + yz = 1*1 + 2*1 = 3
In this case, \(\frac{w}{x} + \frac{y}{z} = \frac{1}{1} + \frac{2}{1} = 3\), so the answer to the question stem is YES.
INSUFFICIENT.
Statement 2: wz + xy = ODD
Consider the following cases:
15 is an integer and \(\frac{15}{3}\) is an integer, so \(\frac{15}{3}\) is a FACTOR OF 15.
21 is an integer and \(\frac{21}{7}\)is an integer, so \(\frac{21}{7}\) is a FACTOR OF 21.
Using the same reasoning:
wz + xy is an integer and \(\frac{wz + xy}{xz}\) is an integer, so \(\frac{wz + xy}{xz}\) is a FACTOR of wz + xy.
Since wz + xy is odd, all of its factors must be odd.
Thus,\(\frac{wz + xy}{xz}\) must be odd.
Since \(\frac{wz + xy}{xz} = \frac{w}{x} + \frac{y}{z}\) = ODD, the answer to the question stem is YES.