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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]

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31 Jan 2013, 02:31

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]

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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

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.

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

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10 Apr 2015, 07:01

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

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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).

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