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What is the greatest common divisor of positive integers m
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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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Lahoosaher
Originally posted by lahoosaher on 07 Jun 2009, 10:12.
Last edited by Bunuel on 28 Dec 2013, 02:51, edited 2 times in total.
Edited the question and added the OA




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Re: What is the greatest common divisor of the positive integers
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02 Mar 2012, 11:20
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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07 Jun 2009, 17:29
amolsk11 wrote: what is the greatest common divisor of the positive integers m and n. 1)m is prime 2)2n=7m m&n GCF? 1) m is prime no info on n, insuff 2) 2n=7m; n=7/2 m or 1:3.5 ratio n could be 2, m could be 7, GCF=1 n could be 8, m could be 28, GCF=4 nsuff together m has to be 7 for n to be an integer, GCF=1



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Re: DS : GCD
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04 Jul 2011, 02:50
option 2 gives 2n=7m. Therefore n=7m/2 HCF of m and 7m/2 is always m/2. So answer should be optin 2 right. what do you think guys



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Re: what is the greatest common divisor of the positive integers
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02 Mar 2012, 10:48
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
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24 Sep 2012, 21:02
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.
So answer is C.
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Re: What is the greatest common divisor of the positive integers
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24 Sep 2013, 10:07
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 the positive integers
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24 Sep 2013, 14:12
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 the positive integers
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24 Sep 2013, 22:12
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 the positive integers
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Re: What is the greatest common divisor of the positive integers
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25 Sep 2013, 00:26
Bunuel wrote: 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. Got it, may be I am getting panic as my exam is coming closer, so could not see even clear things.
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Re: What is the greatest common divisor of positive integers m
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03 Oct 2016, 08:19
I marked B for this question.
My reasoning is  2n = 7m Since 2 and 7 don't have any common factors other than 1, 'm' must be a multiple of 2 and 'n' a multiple of 7. So GCD(m,n) = m/2 = n/7
My doubt is, in such questions are we not allowed to deduce an answer in terms of variables 'm' and 'n'?



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Re: What is the greatest common divisor of positive integers m
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03 Oct 2016, 08:52
puchku wrote: I marked B for this question.
My reasoning is  2n = 7m Since 2 and 7 don't have any common factors other than 1, 'm' must be a multiple of 2 and 'n' a multiple of 7. So GCD(m,n) = m/2 = n/7
My doubt is, in such questions are we not allowed to deduce an answer in terms of variables 'm' and 'n'? Try taking the values of m = 2,4,6 and as we know n = 7m/2, we can have n = 7,14,21, and so on. When combined with Statement 1, we can say m could be only 2. Thus n could be only 7. Hence, HCF= 1.
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Re: What is the greatest common divisor of positive integers m
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03 Oct 2016, 11:14
abhimahna wrote: puchku wrote: I marked B for this question.
My reasoning is  2n = 7m Since 2 and 7 don't have any common factors other than 1, 'm' must be a multiple of 2 and 'n' a multiple of 7. So GCD(m,n) = m/2 = n/7
My doubt is, in such questions are we not allowed to deduce an answer in terms of variables 'm' and 'n'? Try taking the values of m = 2,4,6 and as we know n = 7m/2, we can have n = 7,14,21, and so on. When combined with Statement 1, we can say m could be only 2. Thus n could be only 7. Hence, HCF= 1. But even without considering Statement 1, on the basis of Statement 2 we can say that GCM(m.n) will be m/2 For example m=2 => n=7 => GCD(2,7) = m/2 = 1 m=4 => n=14 => GCD(4,14) = m/2 = 2 Now, even though the actual values are different here, can't we assume that we know the values of 'm' and 'n' (since we want to find their GCD), and thus, in turn, we know the value of GCD which is m/2 But since this is an official question and the answer is C, I am guessing that in DS questions, we have to be able to determine exact values and not such relations



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Re: What is the greatest common divisor of positive integers m
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04 Oct 2016, 08:58
puchku wrote: abhimahna wrote: puchku wrote: I marked B for this question.
My reasoning is  2n = 7m Since 2 and 7 don't have any common factors other than 1, 'm' must be a multiple of 2 and 'n' a multiple of 7. So GCD(m,n) = m/2 = n/7
My doubt is, in such questions are we not allowed to deduce an answer in terms of variables 'm' and 'n'? Try taking the values of m = 2,4,6 and as we know n = 7m/2, we can have n = 7,14,21, and so on. When combined with Statement 1, we can say m could be only 2. Thus n could be only 7. Hence, HCF= 1. But even without considering Statement 1, on the basis of Statement 2 we can say that GCM(m.n) will be m/2 For example m=2 => n=7 => GCD(2,7) = m/2 = 1 m=4 => n=14 => GCD(4,14) = m/2 = 2 Now, even though the actual values are different here, can't we assume that we know the values of 'm' and 'n' (since we want to find their GCD), and thus, in turn, we know the value of GCD which is m/2 But since this is an official question and the answer is C, I am guessing that in DS questions, we have to be able to determine exact values and not such relationsHighlighted line above is the rule for DS questions. We should have a definite answer to arrive at any conclusion.
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Re: What is the greatest common divisor of positive integers m
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26 Jan 2017, 11:07
(1) No info about n but we could test below values as well that will make this statement insufficient. if m = 3 (which is prime) and n = 6, then the gcf is 3. if m = 3 (which is prime) and n = 5, then the gcf is 1. insufficient. (2)In the case of this statement, you can divide by 2m on both sides, to give n/m = 7/2. (you could also divide by 7n, to give m/n = 2/7.) so, the ratio of n to m is 7:2. if they're actually 7 and 2, the gcf is 1. if they're multiples of these numbers, then the gcf is not 1. (for instance, if they're 14 and 4, the gcf is 2.) insufficient.  (together) if you need a prime, and the ratio is 7 to 2, then the numbers must actually be 7 and 2. sufficient.
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What is the greatest common divisor of positive integers m
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15 Jul 2017, 22:55
What is the greatest common divisor of positive integers m and n.
(1) m is a prime number (2) 2n=7m
Statement 1: m is prime. Lots of numbers are prime. INSUFFICIENT.
Statement 2: 2n = 7m
n is a multiple of 7, m is a multiple of 2.
n could be 7 and m could be 2. In that case the GCF is 1. n could be 35 and m could be 10, in which case the GCF is going to be 5. INSUFFICIENT.
Statements 1 & 2:
If m has to be prime, the only prime multiple of 2 is 2. So m = 2 and n = 7. Therefore the GCF 1 is.
(C)



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Re: What is the greatest common divisor of positive integers m
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17 Jul 2017, 19:55
lahoosaher wrote: What is the greatest common divisor of positive integers m and n.
(1) m is a prime number (2) 2n=7m St 1 Obv insuff no info about n St 2 n =7m/2  which must an integer so m must be a multiple of "2" and n must be a multiple of 7 but this could mean several values such as 2, 4 ,8 St 1 and St 2 The only prime number that is a multiple of 2 is 2 so n must be 7 and m must be 2 C



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Re: What is the greatest common divisor of positive integers m
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30 Jul 2017, 16:35
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
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15 Oct 2017, 22:56
lahoosaher wrote: What is the greatest common divisor of positive integers m and n.
(1) m is a prime number (2) 2n=7m each 1 and 2 are not sufficient. use both 1 and 2 m=2n/7 m must be interger, so n must be divided by 7. if n/7 is more than 2, m can not be price. so n must be 7 m must be 2.




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