Hades wrote:

If \(Y\) is a factor of \(Z\), is \(X\) even?

(1) \(\frac{X^{3}Z}{Y} + X\) is even

(2) \(XY\) is even

Question:\((X\) even\()\)?

(1) \(\frac{X^{3}Z}{Y} + X\) is even

\(\longrightarrow X(\frac{X^{2}Z}{Y} + 1)\) is even

\(\longrightarrow X\) even OR \(\frac{X^{2}Z}{Y} + 1\) even

If \(X\) even then we have a

YES, but if \(X\) not even then:

\(\longrightarrow \frac{X^{2}Z}{Y} + 1\) even

\(\longrightarrow \frac{X^{2}Z}{Y}\) odd

\(\longrightarrow \frac{XZ}{Y}\) odd

\(\longrightarrow X odd & \frac{Z}{Y}\) odd

(2) \(XY\) even

\(\longrightarrow (X\) even, \(Y\) odd) or (\(Y\) even, \(X\) odd) or (\(X & Y\) even)

So we have a YES case & a NO case \(\longrightarrow INSUFFICIENT\)

(1&2) Combining

Look at (1) and look at the case \(X odd\). From (2) we know that \(Y even\). But (1) can still be true because \(Z=MY\) where \(M odd\) and \(Y even\), and everything holds.

NO case (X odd)ie \(X=1\), \(Y=2\), \(Z=6\).

\(XY = 2\) (even-- 2 is satsified)

\(\frac{X^{3}Z}{Y} + X = \frac{1^{3}6}{2} + 1 = \frac{6}{2} + 1 = 3 + 1 = 4\) (even)

YES case (X even)ie \(X=2\), \(Y=2\), \(Z=6\).

\(XY = 4\) (even-- 2 is satsified)

\(\frac{X^{3}Z}{Y} + X = \frac{2^{3}6}{2} + 2 = \frac{48}{2} + 2 = 24 + 2 = 26\) (even)

YES & NO\(\longrightarrow INSUFFICIENT\)

\(\longrightarrow INSUFFICIENT\)

Final Answer, \(E\).

_________________

Hades