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What is the greatest common divisor of positive integers m a

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

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: what-is-the-greatest-common-divisor-of-positive-integers-m-79412.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 28 Dec 2013, 03:50, edited 1 time in total.
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Re: gcd Of Prime and another Int [#permalink]

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msunny wrote:
Attachment:
gcdOfPrimeandanotherInt.GIF


[Reveal] Spoiler:
C

1. If m = n, then its either m or n. If m and n are different then it could be 1, m, or 2m, 3m, 4m, 5m, and so on... NSF.

2. If 2n = 7m, n could be a multiple of 7 and m could be a multiple of 2.
If n = 7 and m = 2, then it is 1.
If n = 10 and m = 35, then it is 5.

NSF..

1 and 2: If m is a prime and 2n = 7m, then m must be 2 and n must be 7. So its 1. SUFF...

C.
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Re: gcd Of Prime and another Int [#permalink]

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New post 14 Dec 2009, 02:03
2nd itself is not suff as
2n=7m---> n can be a multiple of & and m can be a multiple of 2

m,n=7,2;4,14;6,21...
gcd differs for each pair

from bth m can have only one value i.e 2---->n=7 and gcd is 1
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Re: gcd Of Prime and another Int [#permalink]

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New post 14 Dec 2009, 08:53
Nice explanation. Thanks.

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Re: gcd Of Prime and another Int [#permalink]

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New post 14 Dec 2009, 16:16
thanks gmat tiger .. kudos +1
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What is the greatest common divisor of positive integers m [#permalink]

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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

Last edited by Bunuel on 02 Mar 2012, 12:22, edited 1 time in total.
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Re: Numbers prob [#permalink]

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New post 20 Dec 2010, 00:13
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: Numbers prob [#permalink]

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New post 20 Dec 2010, 00:24
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.

Hope it clears.
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Re: Which is the greatest common divisor the two positive [#permalink]

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New post 27 Jan 2013, 05:09
mjg2110 wrote:
Hi everybody!

I have some difficulties with this question.

Which is the greatest common divisor the two positive integers m and n?

1) m is a prime
2) 2n=7m

Thanks!


1) m can take several values and there is no information about n. Insufficient.
eg : m = 2, n = 1, GCF = 1
m = 2, n = 2, GCF = 2

2) m and n can take several values. Insufficient.
eg : m = 2, n = 7, GCF = 1
m = 4, n = 14, GCF = 2

1 & 2 together,

\(m = \frac{2}{7}n\). So, n can only be 7 because any other value of n will either give a fraction or a non prime value for m.

Hence values of n & m are known. Sufficient.
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Re: What is the greatest common divisor of positive integers m [#permalink]

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New post 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.

gcd(m,n) = gcd(2,7) = 1.

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

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New post 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

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Re: gcd Of Prime and another Int [#permalink]

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New post 27 Dec 2013, 16:58
GMAT TIGER wrote:
msunny wrote:
Attachment:
gcdOfPrimeandanotherInt.GIF


[Reveal] Spoiler:
C

1. If m = n, then its either m or n. If m and n are different then it could be 1, m, or 2m, 3m, 4m, 5m, and so on... NSF.

2. If 2n = 7m, n could be a multiple of 7 and m could be a multiple of 2.
If n = 7 and m = 2, then it is 1.
If n = 10 and m = 35, then it is 5.

NSF..

1 and 2: If m is a prime and 2n = 7m, then m must be 2 and n must be 7. So its 1. SUFF...

C.


You need to switch "n" and "m" in (2).

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

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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: what-is-the-greatest-common-divisor-of-positive-integers-m-79412.html
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Re: What is the greatest common divisor of positive integers m a [#permalink]

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

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Statement 1: m is a prime.

First thought might be that any number would share only 1 as the GCD. I did so. I took a prime e.g. 39.

What if the two numbers are 39 and 78. In that case, the GCD would be 39. Hence, Insuff

Statement 2.

2n = 7m ----> n = (7/2)m

Substitute even numbers starting from 0. Multiple values again. Insuff..

Combining the 2.

Only m=2 fits in the requirements of both the statement. Hence C
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Re: What is the greatest common divisor of positive integers m   [#permalink] 10 Jun 2016, 09:48
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