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

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If x and y are positive integers such that x = 8y + 12, what [#permalink] New post 17 Sep 2010, 03:23
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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, where u is an integer.

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

well I guess the first questions was quite easy. How about this one? do you still use numbers to solve?
[Reveal] Spoiler: OA

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Re: GCD 2 (Tougher) [#permalink] New post 17 Sep 2010, 03:26
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rafi 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) y = 12z, where z is an integer.

well I guess the first questions was quite easy. How about this one? do you still use numbers to solve?


OK. Algebraic approach:

Given: x=8y+12.

(1) x=12u --> 12u=8y+12 --> 3(u-1)=2y --> the only thing we know from this is that 3 is a factor of y. Is it GCD of x and y? Not clear: if x=36, then y=3 and GCD(x,y)=3 but if x=60, then y=6 and GCD(x,y)=6 --> two different answers. Not sufficient.

(2) y=12z --> x=8*12z+12 --> x=12(8z+1) --> so 12 is a factor both x and y.

Is it GCD of x and y? Why can not it be more than 12, for example 13, 16, 24, ... We see that factors of x are 12 and 8z+1: so if 8z+1 has some factor >1 common with z then GCD of x and y will be more than 12 (for example if z and 8z+1 are multiples of 5 then x would be multiple of 12*5=60 and y also would be multiple of 12*5=60, so GCD of x and y would be more than 12). But z and 8z+1 CAN NOT share any common factor >1, as 8z+1 is a multiple of z plus 1, so no factor of z will divide 8z+1 evenly, which means that GCD of x and y can not be more than 12. GCD(x,y)=12. Sufficient.

Answer: B.

Hope it's clear.
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Re: GCD 2 (Tougher) [#permalink] New post 17 Sep 2010, 03:30
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...
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Re: GCD 2 (Tougher) [#permalink] New post 17 Sep 2010, 03:33
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An easy way to show that z and 8z+1 have no common factors is to use :

gcd(a,b) = gcd(a-b,b) when a>b


So gcd(z,8z+1)=gcd(z,1)=1
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Re: GCD 2 (Tougher) [#permalink] New post 17 Sep 2010, 03:35
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rafi wrote:
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...


There is a general rule: if a and b are multiples of k and are k units apart from each other then k is greatest common divisor of a and b.

For example if a and b are multiples of 7 and a=b+7 then 7 is GCD of a and b.

So if we apply this rule to (2) we would have: both x and y are multiple of 12 and are 12 apart each other, so 12 is GCD of x and y.

So in my previous post I just showed the way this general rule is derived.

Hope it helps.
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Re: GCD 2 (Tougher) [#permalink] New post 17 Sep 2010, 09:49
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Bunuel wrote:
rafi wrote:
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...


There is a general rule: if a and b are multiples of k and are k units apart from each other then k is greatest common divisor of a and b.

For example if a and b are multiples of 7 and a=b+7 then 7 is GCD of a and b.

So if we apply this rule to (2) we would have: both x and y are multiple of 12 and are 12 apart each other, so 12 is GCD of x and y.

So in my previous post I just showed the way this general rule is derived.

Hope it helps.



wow Thanx Bunuel!!
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Re: GCD 2 (Tougher) [#permalink] New post 19 Sep 2010, 06:26
shrouded1 wrote:
An easy way to show that z and 8z+1 have no common factors is to use :

gcd(a,b) = gcd(a-b,b) when a>b


So gcd(z,8z+1)=gcd(z,1)=1


Cool! I'll definitely use that! thanks!
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GMAT Prep - set of questions [#permalink] New post 25 Oct 2010, 06:36
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Re: GCD 2 (Tougher) [#permalink] New post 25 Oct 2010, 07:55
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rafi: Same logic as that given by Bunuel and shrouded1 above, just worded differently in case you have come across this before: "Two consecutive integers do not have any common factors other than 1"

So 8z and 8z + 1 will not share any factors other than 1 and all factors of z will be factors of 8z too. Therefore, z and 8z + 1 will not have any common factors other than 1.
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Re: GCD 2 (Tougher) [#permalink] New post 19 Nov 2013, 00:24
Bunuel wrote:
rafi wrote:
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...


There is a general rule: if a and b are multiples of k and are k units apart from each other then k is greatest common divisor of a and b.

For example if a and b are multiples of 7 and a=b+7 then 7 is GCD of a and b.

So if we apply this rule to (2) we would have: both x and y are multiple of 12 and are 12 apart each other, so 12 is GCD of x and y.

So in my previous post I just showed the way this general rule is derived.

Hope it helps.


Bunuel I have a question with statement 2. Kindly clarify:

if y= 12z and x= 8*12z+12 , then x - y = 96z + 12 - 12z = 84z + 12.

84z + 12 is not euqal to 12

how is it that you're saying x and y are 12 units apart from each other?

Thank you.
Re: GCD 2 (Tougher)   [#permalink] 19 Nov 2013, 00:24
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