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1) no info about n (insuf) 2) n=7m/2. To make n an integer, m=2b (b is an pos integer). With b=1,2,3...we have varied GCD of m and n (insuf) Together we have m=2, n=7 GCD=1, suf

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 [#permalink]

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24 Sep 2012, 22:02

1) This statement says that M is Prime no, So N can be Prime/Composite. If N is Prime , clearly GCD will be 1, If N is composite also GCD will be 1( Except when M itself is a divisor of N, means N<>kM(not equals)), If N=kM then GCD(M,N) will be M it self.(where k is an integer)

2)2N=7M, its clearly not sufficient.

Combining:

From the statement 1, if we can get N=kM or not(where k is an integer) then we will be sure whats the GCD. As from the statement 2, we can see that N=7/2 M, and 7/2 is not an integer. So clearly GCD will be 1.

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

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24 Sep 2013, 11:07

Bunuel wrote:

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.

Answer: C.

Greatest Common divisor and Highest common factor are same thing Bunuel?

Because n= 7m/2 (Taking both this is true only for m = 2) So Greatest common divisor is 2 not 1, Isn't it?
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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.

Answer: C.

Greatest Common divisor and Highest common factor are same thing Bunuel?

Because n= 7m/2 (Taking both this is true only for m = 2) So Greatest common divisor is 2 not 1, Isn't it?

Yes, GCD and GCF are the same thing.

But couldn't understand your second point: the greatest common divisor of 2 and 7 is 1. How can it be 2? Is 7 divisible by 2?
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Re: What is the greatest common divisor of positive integers m [#permalink]

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16 Nov 2014, 07:14

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

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27 Nov 2015, 01:46

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