GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Oct 2019, 11:07 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  S95-12

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58396

Show Tags 00:00

Difficulty:   55% (hard)

Question Stats: 55% (01:15) correct 45% (01:08) wrong based on 82 sessions

HideShow timer Statistics

Is the positive integer $$n$$ divisible by $$6$$?

(1) $$\frac{n^2}{180}$$ is an integer.

(2) $$\frac{144}{n^2}$$ is an integer.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58396

Show Tags

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.

_________________
Intern  Joined: 26 Oct 2015
Posts: 1

Show Tags

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58396

Show Tags

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.
_________________
Intern  Joined: 23 Jan 2018
Posts: 1

Show Tags

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   [#permalink] 23 Jan 2018, 08:55
Display posts from previous: Sort by

S95-12

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  