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:

What is the greatest common divisor of positive integers m and n?

(1) m is a prime number --> if \(m=2=prime\) and \(n=1\) then \(GCD(m,n)=1\) but if \(m=2=prime\) and \(n=4\) then \(GCD(m,n)=2\). Two different answers, hence not sufficient.

(2) 2n=7m --> \(\frac{m}{n}=\frac{2}{7}\) --> \(m\) is a multiple of 2 and \(n\) is a multiple of 7, but this is still not sufficient: if \(m=2\) and \(n=7\) then \(GCD(m,n)=1\) (as both are primes) but if \(m=4\) and \(n=14\) then \(GCD(m,n)=2\) (basically as \(\frac{m}{n}=\frac{2x}{7x}\) then as 2 and 7 are primes then \(GCD(m, n)=x\)). Two different answers, hence not sufficient.

(1)+(2) Since from (1) \(m=prime\) and from (2) \(\frac{m}{n}=\frac{2}{7}\) then \(m=2=prime\) and \(n=7\), hence \(GCD(m,n)=1\). Sufficient.

From stmt 1 - since M is a prime number, and we do not have any info about n, we cannot say anything, hence insuff. From stmt 2 - 2n = 7m. This statement does not say anything about m and n. It only says that m/n = 2/7 . The number could be anything {2,7} or {6, 21} . Both the cases produce different highest common divisor. So insuff.

Taking both the stmts together - what we know/deduce is - Divisors of product of two prime will be 1, the prime number1, the prime number 2, and the product of two prime num. so for 7m = {1, 7, m , 7m} and for 2n = 7m, given m to be a prime number, m has to be 2. If m is 2, then n = 7 and hence suff.

Re: What is the greatest common divisor of positive integers m [#permalink]

Show Tags

31 Jan 2013, 02:31

Expert's post

We have n and m as positive integers.

From F.S 1, we have m is a prime.Let us assume m=7. Thus, for n=14, we have gcd(m,n) as 7, for n=6 we have gcd(m,n) as 1. Thus this statement by itself is not sufficient.

From F.S 2, we have 2n=7m. Thus, n = 7m/2. Now as n,m are integers, m=2k(k is an integer). Thus, we get

n=7k, m=2k. As both 2 and 7 are prime, the gcd(m,n) here will be k, and this can have any value(1,2,3...);Thus not sufficient.

Combining both the F.S, we know m is prime and m=2k. Thus k can not be anything except 1, else m won't be a prime anymore. Thus, k=1 and n=7,m=2.

Re: What is the greatest common divisor of positive integers m [#permalink]

Show Tags

15 Feb 2013, 18:44

jullysabat wrote:

What is the greatest common divisor of positive integers m and n ?

(1) m is a prime number

(2) 2n = 7m

What is the GCF of m & n?

(1) Insufficient- it can be any set of #'s (2 & 6, 3 & 12) (2) Insufficient- you can plug in any #'s that make the equation equal (n=35 & m=10 - GCF is 5 or n=26 & m=8 - GCF is 2)

You know m is prime so the only way to balance out the equation is to replace m as 2 (only even prime #) b/c whatever 2n produces it will be an even #.

odd * odd = odd even * even = even even * odd = even

Re: What is the greatest common divisor of positive integers m [#permalink]

Show Tags

10 Jun 2016, 09:48

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. _________________

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

Cal Newport is a computer science professor at GeorgeTown University, author, blogger and is obsessed with productivity. He writes on this topic in his popular Study Hacks blog. I was...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...