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If the greatest common factor of two integers, m and n, is

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If the greatest common factor of two integers, m and n, is [#permalink] New post 20 Mar 2011, 08:52
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If the greatest common factor of two integers, m and n, is 56 and the least common multiple is 840, what is the sum of the m and n?

(1) m is not divisible by 15.
(2) n is divisible by 15.
[Reveal] Spoiler: OA

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Re: Number Prop DS [#permalink] New post 05 Sep 2011, 11:21
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Quote:
If the greatest common factor of two integers, m and n, is 56 and the least common multiple is 840, what is the sum of the m and n?

(1) m is not divisible by 15.
(2) n is divisible by 15.


prime factors of 56: 7, 2, 2, 2
prime factors of 840: 7, 2, 2, 2, 3, 5

From Statement 1
m = 56*3 or m=56
Insufficient

From Statement 2

m=56; n=840
sufficient

Answer: B
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Re: Number Prop DS [#permalink] New post 20 Mar 2011, 09:02
s1 insufficient
Consider m=56*3 and n=56*5
consider m = 56 and n=56*15

s2 sufficient
m=56 and n=56*15

Hence B

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Re: Number Prop DS [#permalink] New post 20 Mar 2011, 10:22
1 not sufficient
as there more than one possible combination for m and n

m = (2^3)7(3) n = (2^3)7(5)

m = (2^3)7 n = (2^3)7(15)

2. Sufficient

only possible combination for m and n here is m = (2^3)7 n = (2^3)15

Hence answer is B.
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Re: Number Prop DS [#permalink] New post 20 Mar 2011, 19:12
56 = 2^3 * 7

So m and n have 2^3 * 7 as factor

840 = 7 * 120 = 7 * 5 * 24 = 7 * 5 * 3 * 2^3

(1), m is m is not divisible by 15, so m does not have 5 and 3 as factor


So m = 2^3 * 7 * k (where k is an intger other than 3 or 5)


Now m*n = 56 * 840

So n = 56/56k * 840 = 840/k , which is not sufficient as n could be 840, or 840/56 = 15


(2) n is divisible by 15, so n has 3 and 5 as factor

So n = 3*5* 2^3 * 7*p, where p is an integer

=> m = 56*840/15*56p = 56*56/56p, so m can be 56/p, now m has to be minimum 56, so p = 1, hence m = 56 and n = 840

So answer is B
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Re: Number Prop DS [#permalink] New post 21 Mar 2011, 21:10
Product of M & N= LCM *GCD
==>56*840
==>7^2*2^6*5*3

Stmt 1 m could be 7*2^3*5 or 7*2^3*3..
Stmt 2
Since GCD is 56 both m and n should have 7 *2^3
n =>7 *2^3 *3*5 (n is divisible by 15 so it should have 3 and 5 as a factor).. Sufficient

B
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Re: If the greatest common factor of two integers, m and n, is [#permalink] New post 15 Oct 2013, 16:27
It is given GCF = 56 = 7 x 2 x 2 x 2
and LCM = 840 = 56 (GCF) x 15
For more fundamental elaboration:-
GCF and LCM
----------
7 |m , n
2 |m1, n1
2 |m2, n2
2 |m3, n3
--- 1 , 15
or
--- 3 , 5
From Statement 1 informs "m" is not divisible by 15, so in above illustration, we can have either 1 or 3 under "m", which makes the statement insufficient to identify the value of m,

From Statement 2 informs "n" is divisible by 15, so in above graphic illustration, we can establish that we will have 1 under "m" and 15 under "n", which is sufficient to derive both the value of n and m

The value of m = 1 x 2 x 2 x 2 x 7 = 56
The value of n = 15 x 2 x 2 x 2 x 7 = 840
m + n = 896
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Re: Number Prop DS [#permalink] New post 16 Nov 2013, 14:00
gmatopoeia wrote:
Quote:

From Statement 1
m = 56*3 or m=56
Insufficient
[u]

Answer: B



For the sake of my comprehension, should m=56*5 be a possibility as well?
Re: Number Prop DS   [#permalink] 16 Nov 2013, 14:00
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