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16 Sep 2014, 01:49
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
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Question Stats:
57% (01:39) correct
43%
(01:40)
wrong
based on 65
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Is the positive integer \(n\) divisible by \(6\)?
(1) \(\frac{n^2}{180}\) is an integer.
(2) \(\frac{144}{n^2}\) is an integer.
(1) \(\frac{n^2}{180}\) is an integer.
(2) \(\frac{144}{n^2}\) is an integer.
16 Sep 2014, 01:49
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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
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
29 Oct 2015, 18:10
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Bunuel
Thanks Bunuel, for the first statement, n^2/180. the square root of n has 6 as it factor, but also has a square root of 5. Does it still mean that it can be divisible by 6?
