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# If a and b are positive integers divisible by 6, is 6 the

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Intern
Joined: 16 May 2008
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If a and b are positive integers divisible by 6, is 6 the [#permalink]

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22 Aug 2008, 09:23
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If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

(1) a = 2b + 6

(2) a = 3b
SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York
Re: Tough DS- Greatest common Divisor [#permalink]

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22 Aug 2008, 09:35
fatimoo wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

(1) a = 2b + 6

(2) a = 3b

say b=6m
(1) a= 2b + 6
a= 2 6m +6 = 6*(2m+1)
common greatest divsor of 6*m and 6 *(2m+1) is 6.
m and 2m+1 don't have any common divisors other than 1.
e.g m=1 2m+1 =3
m=2 2m+1 =5

Sufficient
(2) a = 3b
6m and 3*6m
when m=3
6*3 3*6*6 GREATEST COMMON DIVISOR 3*6
WHEN M=1
6 3*6 GREATEST COMMON DIVISOR 6

So.. INSUFFICINET.

A
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Last edited by x2suresh on 22 Aug 2008, 09:54, edited 1 time in total.
Director
Joined: 12 Jul 2008
Posts: 513
Schools: Wharton
Re: Tough DS- Greatest common Divisor [#permalink]

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22 Aug 2008, 09:48
fatimoo wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

(1) a = 2b + 6

(2) a = 3b

A

(1) Sufficient

let b = 6k

a = 2(6k)+6 = 6(2k + 1)
b= 6k

2k+1 and k will only have 1 as a common divisor, so 6 (and its prime components) is the only common factors of a and b.

(2) Insufficient

let b = 6k

a = 3(6k)
b = 6k

Here, 6k (and its components) is common to both, where k could be any integer.
Intern
Joined: 16 May 2008
Posts: 2
Re: Tough DS- Greatest common Divisor [#permalink]

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24 Aug 2008, 13:12
OA is A. Your explanation is much easier than below..

(1) SUFFICIENT: We are already told in the question stem that 6 is a divisor of both a and b. This statement tells us that a is exactly 6 more than 2b. If one number is x units away from another number, and x is also a factor of both of those numbers, than x is also the GCF of those two numbers. This always holds true because x is the greatest number separating the two; in order to have a larger GCF, the two numbers would have to be further apart.

This statement, then, tells us that the GCF of a and 2b is 6. The GCF of a and b can't be larger than the GCF of a and 2b, because b is smaller than 2b; since we were already told that 6 is a factor of b, the GCF of a and b must be also be 6.

This can also be tested with real numbers. If b = 6, then a would be 18 and the GCF would be 6. If b = 12, then a would be 30 and the GCF would be 6. If b = 18, then a would be 42 and the GCF would still be 6 (and so on).

(2) INSUFFICIENT: There are no mathematical rules demonstrated in this statement to help us determine whether 6 is the GCF of a and b. This can also be tested with real numbers. If b = 6, then a would be 18 and the GCF would be 6. If, however, b = 12, then a would be 36 and the GCF would be 12.

Re: Tough DS- Greatest common Divisor   [#permalink] 24 Aug 2008, 13:12
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