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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 [#permalink]

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07 Jun 2009, 10:12
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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
[Reveal] Spoiler: OA

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Lahoosaher

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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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
[Reveal] Spoiler:
C

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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08 Jun 2009, 18:14
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
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04 Jul 2011, 02:50
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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 [#permalink]

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02 Mar 2012, 10:48
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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 the positive integers [#permalink]

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02 Mar 2012, 11:20
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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.

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

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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.

---If it is stupid but it works, it isn't stupid.
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Re: What is the greatest common divisor of the positive integers [#permalink]

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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.

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 [#permalink]

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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.

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 [#permalink]

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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 [#permalink]

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24 Sep 2013, 23:54
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 the positive integers [#permalink]

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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 [#permalink]

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16 Nov 2014, 06:14
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27 Nov 2015, 00:46
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Re: What is the greatest common divisor of positive integers m [#permalink]

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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 [#permalink]

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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 [#permalink]

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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 [#permalink]

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04 Oct 2016, 07:54
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 relations

Can some expert please shed light on my interpretation?

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

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04 Oct 2016, 08:58
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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 relations

Highlighted 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 [#permalink]

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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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Re: What is the greatest common divisor of positive integers m   [#permalink] 26 Jan 2017, 11:07
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