Is the positive integer n divisible by 6?

Official Solution:

We must determine whether \(n\) is evenly divisible by \(6\). In other words, we must determine whether \(n\) has a factor of 2 and a factor of 3 (since \(2 \times 3 = 6\)). If \(n\) is evenly divisible by 6, then \(\frac{n}{6} = k\) for some integer \(k\), and we can rewrite \(n\) as \(6k\).

Statement 1 says that \(\frac{n^2}{180}\) is an integer. In order for \(n\) to be divisible by 6, \(n^2\) must have two factors of 6, since \((6k)^2 = 36k^2\). Since \(n^2\) is divisible by 180, 180 is a factor of \(n^2\), and all factors of 180 are also factors of \(n^2\). Because 180 has 36 as a factor, and \(36 = 6 \times 6\), \(n^2\) has two factors of 6. Thus, \(n\) has 6 as a factor, and \(n\) is divisible by 6. Statement 1 is sufficient to answer the question. Eliminate answer choices B, C, and E. The correct answer choice is either A or D.

Statement 2 says that \(\frac{144}{n^2}\) is an integer. The prime factorization of 144 is \(2 \times 2 \times 2 \times 2 \times 3 \times 3\). This means that \(n^2\) can contain up to four factors of 2 and two factors of 3 (note that any given factor must occur an even number of times in the prime factorization of \(n^2\)). If \(n = 4\), the statement is satisfied, since \(\frac{144}{n^2} = \frac{144}{4^2} = \frac{144}{16} = 9\). In this case, \(n\) is NOT divisible by 6. However, \(n\) could also equal 6, since \(\frac{144}{n^2} = \frac{144}{6^2} = \frac{144}{36} = 4\). In this case, \(n\) is divisible by 6. Statement 2 does NOT provide sufficient information to answer the question.


Answer: A.