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 Your Progress
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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
I've used just common sense here, since I'm not getting anywhere with traditional ways.
(A) St 1 is sufficient. Because, since 18 = 3*3*2, 5 and 18 have no factors in common.
Which means for 5n/18 to be an integer, n must be a multiple of 18!
Hence sufficient.
(B) is not sufficient because 3 and 18 have common factors, so n may be a multiple of 18, the only thing certain here is that it will be a multiple of 6. Therefore not sufficient.
(1) If 5n/18 is an integer, call it k, we know that n/18=k/5, which may or may not be an integer. NOT SUFF
If (5n/18)=k=>5n=18k=>n=(18k/5) since both n and k are INTEGERS then k is multiple of 5 and respectively n is a multiple of 18 and an INTEGER so A) is SUFF byitself,
lets see..the stem says nothing about N being an integer...soo
1) 5 N/18 - integer...well N can be an integer multiple of 18 or it can be 18M/5 where M is an integer not divisible by 5...Insuff
2) simplify this to N/6, so N has prime factors 2 and 3, we still dont know if it has enuff 3s or not...but we noe for sure that N is not a fraction...
combining them we now now that N is not a fraction and that its a mutliple of 18....
C it is...
Answer woud be A if we are told in the stem that N is an integer..
Re: Is n/18 an integer? (1) 5n/18 is an integer. (2) 3n/18 is an [#permalink]
18 Nov 2013, 03:54
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
5n/18=integer If we consider n=18M/5 Where M can be number which is not divisible by 5 .. If we consider this case, then 5(18M/5)/18=integer is not fulfilling sinceM is not integer , to fulfill this M must be integer hence N is integer..
Re: Is n/18 an integer? [#permalink]
30 Nov 2013, 23:30
2
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
shinewine wrote:
Is n/18 an integer?
(1) 5n/18 is an integer.
(2) 3n/18 is an integer.
From F.S 1, we know that \(\frac{5n}{18}\) is an integer. For\(\frac{n}{18} = 1\), we have a YES. Again, for \(\frac{n}{18} = \frac{1}{5}\) , we have a NO.Insufficient.
From F.S 2, we know that\(\frac{3n}{18}\) is an integer. For \(\frac{n}{18} = 1\), we have a YES,but for \(\frac{n}{18} = \frac{1}{3}\) , we have a NO.Insufficient.
Taking both together, we know that from F.S 1, either\(\frac{n}{18} = \frac{k}{5}\) or \(\frac{n}{18} = p\) , where k,p are integers and k and 5 are co-primes.
But, for \(\frac{n}{18} = \frac{k}{5}\), \(\frac{3*n}{18} = \frac{3*k}{5}\) and it will not be an integer. Thus, \(\frac{n}{18} = p\) can the only be form possible.
Re: Is n/18 an integer? [#permalink]
03 Dec 2013, 00:07
6
This post received KUDOS
2
This post was BOOKMARKED
shinewine wrote:
Is n/18 an integer?
(1) 5n/18 is an integer.
(2) 3n/18 is an integer.
We need to find whether n is a multiple of 18 or n= 18I for some integer I
St 1: 5n/18= Integer or n = 18*Integer/ 5 Now if Integer = 2, then n =7.2 and n/18 is not an integer
but integer = 5 then n = 18 and 18/18 is an integer ------Possible values of n = 3.6,7.2,10.8,14.4,18......
So A and D ruled out
St 2: we see that n is a multiple of 6 so possible values of n =6,12,18...... If n= 6 then n/18 is not an integer but if n=18 then n/18 is an integer. So option B ruled out
Combining we get possible values of n =18,36,54 and so on
Hence Ans C _________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
Re: Is n/18 an integer? [#permalink]
11 Feb 2014, 11:53
From 1: 5n/18 = a (an integer) => n/18 = a/5 (we cannot be sure if this is an interger) From 2: 3n/18 = b (an integer) => n/18 = b/3 (we cannot be sure if this is an interger) Combining 1&2: 2*(statement 2) - (statement 1) => 6n/18 - 5n/18 = 18*2*b - 18*a => n/18 = 18(2b-a) {this we know for sure is an integer} Thus, Option C
Re: Is n/18 an integer? [#permalink]
13 Feb 2014, 19:12
2
This post received KUDOS
It's C, but I think this is as clear as it can be explained:
S1: \(\frac{5n}{18}=K1\) (Letting K1 be an integer)
Then, isolating n we get \(n=\frac{18K1}{5}\)
Going back to the original question if n is divisible by 18, based on this answer is "Maybe" (provided that K1 is a multiple of 5)
Insufficient.
S2: \(\frac{3n}{18}=K2\) (Letting K2 be an integer)
Then, isolating n we get \(n=\frac{18K2}{3}\), reducing \(n=6K2\)
Going back to the original question if n is divisible by 18, based on this answer is "Maybe" (Provided that K2 is a multiple of 3)
Insufficient.
S1 & S2: We will make the expressions for n equal:
\(\frac{18K1}{5}=6K2\), simplifying \(K2=\frac{3K1}{5}\) The key here is understanding that K1 and K2 MUST be integers. As such, the "Maybes" of S1 and S2 are proven to be true. From the expression above, K2 is a multiple of 3 and K1 is a multiple of 5.
This is perhaps an extended explanation, but I think is clear enough.
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...
Ninety-five percent of the Full-Time Class of 2015 received an offer by three months post-graduation, as reported today by Kellogg’s Career Management Center(CMC). Kellogg also saw an increase...