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# S95-12

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Math Expert
Joined: 02 Sep 2009
Posts: 50004

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16 Sep 2014, 01:49
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Difficulty:

55% (hard)

Question Stats:

55% (01:14) correct 45% (01:08) wrong based on 80 sessions

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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.

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Joined: 02 Sep 2009
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16 Sep 2014, 01:49
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.

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Joined: 26 Oct 2015
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29 Oct 2015, 18:10
Bunuel wrote:
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.

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?
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30 Oct 2015, 01:40
doriazhao wrote:
Bunuel wrote:
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.

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?

Sorry, cannot follow you... Can you please elaborate what you mean? Thank you.
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23 Jan 2018, 08:55
hello,
I think the guy above means that the prime factorizaton of 180 is 2,2,3,3,5 logically from the explanation of (2) follows that it should not work out for 180 either. Would you explain that?
Re: S95-12 &nbs [#permalink] 23 Jan 2018, 08:55
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# S95-12

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