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What is the greatest common divisor of positive integers m and n ?
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Updated on: 06 Jun 2019, 08:23
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What is the greatest common divisor of positive integers m and n ? (1) m is a prime number (2) 2n = 7m
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Originally posted by lahoosaher on 07 Jun 2009, 11:12.
Last edited by Bunuel on 06 Jun 2019, 08:23, edited 3 times in total.
Edited the question and added the OA




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Re: What is the greatest common divisor of positive integers m and n ?
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02 Mar 2012, 12:20
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.
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Re: DS_GCD
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Updated on: 05 Jul 2011, 02:14
What is the greatest common divisor of positive integers M and N ??? 1)M is a prime number 2)2N=7M
If we look at statement 2 and plug in numbers we'll quickly see it's not sufficient.
Let M=2 then N = 7 GCD=1. Let M=6 then N = 21 GCD=3.
S2 basically tells us that 2 is a factor of M and 7 is a factor of N. But we don't know if they have more shared factors or not. Insufficient.
Does that make sense Akshaydiljit?



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Re: What is the greatest common divisor of positive integers m and n ?
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02 Mar 2012, 11:48
OA is C.
1)m is prime Clearly insufficient.
2)2n=7m can be written as n= 7m/2. n & m are integers. Put m=1,2,3,4 .... therefore m has to be a multiple of 2. Insufficient.
Combined m is prime(stat1) and m= multiple of 2(stat2)
Hence m=2 & n=7
GCF is 1.



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Re: What is the greatest common divisor of positive integers m and n ?
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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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Re: What is the greatest common divisor of positive integers m and n ?
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24 Sep 2013, 15:12
honchos wrote: 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? 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 and n ?
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24 Sep 2013, 23:12
Bunuel, Our m is coming as 2, so isn't 2 a GCD, Or may be I have misunderstood the solution?



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Re: What is the greatest common divisor of positive integers m and n ?
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25 Sep 2013, 00:54
honchos wrote: Bunuel, Our m is coming as 2, so isn't 2 a GCD, Or may be I have misunderstood the solution? The question asks: what is the greatest common divisor of positive integers m and n? We got that m=2 and n=7. What is the greatest common divisor of 2 and 7? Is it 2? No, it's 1.
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Re: What is the greatest common divisor of positive integers m and n ?
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30 Jul 2017, 17:35
lahoosaher wrote: What is the greatest common divisor of positive integers m and n.
(1) m is a prime number (2) 2n=7m We need to determine the greatest common divisor, or the greatest common factor (GCF), of integers m and n. Statement One Alone: m is a prime number. Since we don’t know anything about n, statement one is not sufficient to answer the question. We can eliminate answer choices A and D. Statement Two Alone: 2n = 7m We can manipulate the equation 2n = 7m: n = 7m/2 n/m = 7/2 Even with the equation rewritten, we see that there are many options for m and n, and thus there are many different GCFs for m and n. For instance, if n = 7 and m = 2, then the GCF is 1. However, if n = 14 and m = 4, then the GCF is 2. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B. Statements One and Two Together: Using statements one and two, we know that m is prime and that n/m = 7/2. Therefore, m must equal 2 and n must equal 7. When m is 2 and n is 7, the GCF is 1. Answer: C
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Re: What is the greatest common divisor of positive integers m and n ?
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22 Dec 2017, 11:54
lahoosaher wrote: What is the greatest common divisor of positive integers m and n.
(1) m is a prime number (2) 2n=7m Target question: What is the GCD of m and n?Statement 1: m is a prime number If m is a prime number, it has exactly 2 divisors (1 and m), so this tells us that the GCD of m and n must be either 1 or m. Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT. Statement 2: 2n = 7mIf 2n = 7m then we can rearrange the equation to get n = (7/2)m IMPORTANT: Notice that if m were to equal an ODD number, then n would not be an integer. For example, if m = 3, then n = 21/2 (n is not an integer). Similarly, if m = 11, then n = 77/2 (n is not an integer). So, in order for n to be an INTEGER, m must be EVEN.If m must be EVEN, there are several possible values for m and n. Consider these two cases: case a: m = 2 and n = 7, in which case the GCD = 1case b: m = 4 and n = 14, in which case the GCD=2Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT. Statements 1 & 2 combinedFrom statement 1, we know that m is prime, and from statement 2, we know that m is even. Since 2 is the only even prime number, we can conclude that m must equal 2. If m = 2, then n must equal 7, which means that the GCD must be 1. Since we are able to answer the target question with certainty, statements 1 & 2 combined are sufficient, and the answer is C Cheers, Brent
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Re: What is the greatest common divisor of positive integers m and n ?
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23 Nov 2018, 10:13
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. 2) 2n=7m > m/n=2/7 n=3.5m GCF(m,3.5m)= m ? is this correct ? Its not sufficient because we dont have the value of M ?



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Re: What is the greatest common divisor of positive integers m and n ?
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23 Nov 2018, 21:55
renjana wrote: 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. 2) 2n=7m > m/n=2/7 n=3.5m GCF(m,3.5m)= m ? is this correct ? Its not sufficient because we dont have the value of M ? Hello Yes, I think you have concluded properly. GCF of m & 3.5m will depend on the value of m. Eg, if m= 2, then 3.5m = 7, and their GCF will be 1. However, if m= 4, then 3.5m= 14, and their GCF will be 2. So GCF can take multiple values depending on the value of m.



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