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# If x and y are positive integers such that x = 8y +12, what

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Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
If x and y are positive integers such that x = 8y +12, what [#permalink]

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29 Nov 2006, 10:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x and y are positive integers such that x = 8y +12, what is the greatest common divisor of x and y?

(1) x =12u, where u is an integer

(2) x =12z, where z is an integer
_________________

Impossible is nothing

Senior Manager
Joined: 20 Feb 2006
Posts: 373
Re: GMAT Prep: Greatest Common Divisor [#permalink]

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29 Nov 2006, 12:49
Hermione wrote:
If x and y are positive integers such that x = 8y +12, what is the greatest common divisor of x and y?

(1) x =12u, where u is an integer

(2) x =12z, where z is an integer

Both statements essentially tell us the same thing?

From 1) 12u = 8y + 12 or u=2y/3 + 1 u=>3 to give integer values of y INSUFF

2) Same as for 1) INSUFF

Taking both together, x can equal 36 giving GCD of x&y as 3 or 60 giving GCD of 6. INSUFF

E for me
Director
Joined: 06 Feb 2006
Posts: 897

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29 Nov 2006, 13:01
E this one....

Both statements can yield x=12 y=3 or x=36, y=4....
Senior Manager
Joined: 20 Feb 2006
Posts: 373

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29 Nov 2006, 13:14
SimaQ wrote:
E this one....

Both statements can yield x=12 y=3 or x=36, y=4....

If x=12 then y cant be an integer. If x=36 then y=3
Senior Manager
Joined: 01 Oct 2006
Posts: 495

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29 Nov 2006, 14:24
x=8y+12
st1 says x=12u thus 12(u-1)=8y the GCD will change depending upon values of u and y..insuff
st2 same 1 insuff

combining also insuff

Manager
Joined: 25 Nov 2006
Posts: 59

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29 Nov 2006, 15:03
from 1:we can conclude that the value of 'u' may be 3,5,7,9.........
since x,y given to be positive integers.
hence we cannot conclude or get the values of x and y from where we cannot get the gcd.
option E is the wright answer.
Director
Joined: 06 Feb 2006
Posts: 897

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30 Nov 2006, 01:03
SimaQ wrote:
E this one....

Both statements can yield x=12 y=3 or x=36, y=4....

If x=12 then y cant be an integer. If x=36 then y=3

Yes you are right.... was in a rush on this one... however the values do not really matter, the point is that there can be several values....
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK

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30 Nov 2006, 08:07
Oh sh*t! Don't kill me. The second statemt should read

(2) y=12z, where z is an integer

So sorry!!!
_________________

Impossible is nothing

SVP
Joined: 05 Jul 2006
Posts: 1747

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30 Nov 2006, 08:14
If x and y are positive integers such that x = 8y +12, what is the greatest common divisor of x and y?

(1) x =12u, where u is an integer

(2) y =12z, where z is an integer

from one

12(u-1) = 8y.ie: y = 3/2 ( u-1) .insuff

from two

y is a multiple of 12 , from given x = 12*8*y+12 ie a multiple of 12 = 12(8y+1)

thus the greatest common divisor is 12...B is my answer
Manager
Joined: 17 Dec 2004
Posts: 71

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30 Nov 2006, 10:18
I get B.

The stem tells us that x = 8y + 12.

S1: x = 12u, where u is an integer. This tells us that: 12u = 8y + 12 ---> 3u = 2y + 3 ---> 3(u-1) = 2y ---> u = (2y)/3 + 1. Because this can give multiple values for x ---> INSUFF.

S2: y = 12z, where z is an integer. This tells us that: x = 8(12z) + 12 ---> x = 12(8z +1). From this it follows that, for all values of z (and thus y), the greatest common factor is 12 ---> SUFF.

Therefore B.

What is the OA?
Senior Manager
Joined: 20 Feb 2006
Posts: 373

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30 Nov 2006, 15:37
I'm going with B now

x = 8(12z) + 12 ---> x = 96z + 12

GCD = 12

Statement 1) Still INSUFF
2) Now SUFF
Manager
Joined: 01 Feb 2006
Posts: 78
Location: New York

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01 Dec 2006, 07:43
OA is B. you guys got it !!
01 Dec 2006, 07:43
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