Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9.

Is \(x(x-2) > 0\)?

\(x(x-2) > 0\) is true when \(x < 0\) or \(x > 2\). Essentially, if \(x\) is not 0, 1, or 2, we have a YES answer to the question.

(1) \(x\) is a prime number.

If \(x=2\), then the answer is NO, but if \(x\) is some other prime, then the answer is YES. Not sufficient.

(2) \(x^2\) is a multiple of 9.

If \(x=0\), then the answer is NO, but if \(x=3\), then the answer is YES. Not sufficient.

(1)+(2) Since from (1), \(x\) is a prime, and from (2), \(x^2\) is a multiple of 9, then \(x\) can only be 3. Therefore, the answer to the question is YES. Sufficient.

Answer: C.

As for your question: zero is divisible by EVERY integer except zero itself, since 0/integer=0=integer (or, which is the same, zero is a multiple of every integer).