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

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Manager
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If x and y are positive integers such that x=8y+12, what is [#permalink]

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04 Apr 2007, 16:50
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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, u is an integer
(2)y=12z, z is an integer.

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Director
Joined: 14 Jan 2007
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04 Apr 2007, 17:31
Statement1:
x =12u, u an int.
as x is a +ve int, it means u will also be a +ve int.
put the value in equation
x = 8y + 12
==> 12u = 8y + 12 ==> y = 12/8 (u-1)
there is no common factor between u and (u-1)/8. hence GCD of x and y will be 12. SUFF

Statement2:
y = 12z, z an int.
z will also be +ve as y is +ve int.
Put the values in equation
x = 8y +12
==> x = 8*12z + 12 ==> 12(8z-1)
There is no common factor between z and 8z-1, hence GCD of x and y will always be 12. SUFF

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Senior Manager
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04 Apr 2007, 23:39
The OA is already up but...

(1)x=12u, u is an integer.
12u = 8y+12
y = 3,6,9... for
n = 3, 5, 7...
GCD changes as y changes.
If y = 3 then GCD = 3
If y = 6 then GCD = 6
Insuff..

(2)y=12z, z is an integer.
x=8(12z)+12
x=12z(8+1)
x=12z(9)
x=108z
108z is a multiple of 12z therefore the GCD of x and y is 12z.
Suff..

So B.

Is my reasoning correct?

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Director
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04 Apr 2007, 23:59
ricokevin wrote:
.
.
x=8(12z)+12
x=12z(8+1)
.
.

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Senior Manager
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05 Apr 2007, 01:22
Juaz wrote:
ricokevin wrote:
.
.
x=8(12z)+12
x=12z(8+1)
.
.

Oops...

x=8(12z)+12
x=12(8z+1)

so x is a multiple of 12.
y is also a multiple of 12 since y=12z

(8z+1) is always odd so GCD between (8z+1) and z is 1.
Therefore the GCD of x and y is 12?

So B suff?

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05 Apr 2007, 01:22
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