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# gmatprep 2 GCD DS.

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16 Jun 2008, 20:59
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Hi guys have a look at this little nasty GCD problem from gmatprep2.
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Manager
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Re: gmatprep 2 GCD DS. [#permalink]

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16 Jun 2008, 21:11

Statement 1 : x is a multiple of 12 , but y can be anything from 0 to 3 and therefore not possible to determine a single greatest common divisor

Statement 2 : if y is a multiple of 12 then x is also a multiple of 12 , can be found from given equation . So greatest common divisor always 12
CEO
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Re: gmatprep 2 GCD DS. [#permalink]

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16 Jun 2008, 21:34
Is there ever a scenario where GCD could be higher than 12 ??
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Re: gmatprep 2 GCD DS. [#permalink]

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16 Jun 2008, 22:02
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This is DS question #240 of the OG10

Here is the official explanation for statement 2:

statement 2 implies that y is a multiple of of 12 or that 12 is a divisor of y. Since x = 8y + 12, it follows that 12 is a divisor of x, and thus 12 is a common divisor of x and y. Since 12 = x - 8y (rephrase the stem), any common divisor of x and y must be a divisor of 12. Therefore no integer greater than 12 is a common divisor of x and y, and 12 is the GCD of x and y.

this question gets me everytime.
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Re: gmatprep 2 GCD DS. [#permalink]

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16 Jun 2008, 23:05
bsd_lover wrote:
Hi guys have a look at this little nasty GCD problem from gmatprep2.

x = 8y +12
from 1: x = 12u
x = 12u = 8y +12
we know nothing about y. nsf.

2: y = 12z
x = 8 (12z) + 12
x = 12 (8z+1)
now we know x and y are multiple of 12. therefore, 12 is the gcd.
12 is gcd because x can only be a multiple of 12 because no matter the value of z, x is 98z plus 12. so the common divisor of x and y is 12.

for ex:

if z = 1, x = 108 and y = 12
if z = 2, x = 204 and y = 24
if z = 3, x = 300 and y = 36
and so on....................
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Re: gmatprep 2 GCD DS.   [#permalink] 16 Jun 2008, 23:05
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# gmatprep 2 GCD DS.

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