Is n an integer? (1) 3n^2 is an integer.

Official Explanation

We need to know whether n is an integer. We will most likely analyze by cases and by the rules of number properties. On to the data statements, separately first.

Statement (1) tells us that 3n^2 = integer. We can dig into some cases. Say 3n^2 = 18. That's an allowed case, since 18 is an integer. In this case n^2 = 6 and \(n=\sqrt{6}\) , so n is not an integer. We can probably find a case that generates a contradictory answer. Say 3n^2 = 27. That's an allowed case, since 27 is an integer. In this case, n = 3, so n is an integer. Hence we have contradictory answers from allowed cases, so we have insufficient information to answer the question definitively. Statement (1) is insufficient.

Statement (2) tells us that \(\frac{\sqrt{n}}{3}=integer\). That means that

\(\sqrt{n}=3*integer\)

\(\sqrt{n}=integer\)

This is similar to Statement (1), but different. Looking at the right side here: any integer times three yields another integer. So we are left with a (different) integer squared, and any integer squared yields another integer. That means that n will always be an integer. We have sufficient information to answer the question definitively, so Statement (2) is sufficient.

The correct answer is (B).

Video Explanation

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