SAT problem of the week, September 29, 2014 solution

Hint for process of elimination: Though this problem has a “None of the above” option, choosing a random value for $latex z &s=1$ satisfying the simple conditions that $latex z &s=1$ is odd, such as $latex z=3 &s=1$, would immediately eliminate three of the choices because the answers are odd.

A second benefit of choosing a random value of $latex z &s=1$ is that the process may increase your intuition of what the answer should be.

Read on for a full solution.

Answer: C

Solution: I welcome better solutions but here is mine.

If $latex z &s=1$ is odd, then $latex z = 2n+1 &s=1$ for some integer $latex n &s=1$. If we substitute this equation into choice C we get

$latex z^2 – 3z = (2n+1)^2 -3(2n+1) = 4n^2+4n+1 -6n -3 = 4n^2-2n-2 = 2(2n^2 -n-1) &s=1$

The above expression is an integer multiple of 2, so it is even.

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