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

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x and y are positive integers such that x=8y+12, what is the [#permalink]  29 Jan 2012, 17:25
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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.

[Reveal] Spoiler:
For me its B and this is how I solved it. Is my solution correct?

Question is asking for GCD of x and y.

GCF or GCD is the product of common prime factors with lowest exponents. for example GF of 12 and 24 is

12 = 2^2 * 3^1
24 = 2^3 * 3^1

GCF = 2^2 * 3^1 = 12

Coming back to the question and considering statement 1

x is a multiple of 12

So if we put different values of x in the our equation GCF will be different. Therefore this statement is INSUFFICIENT.

Considering statement 2

Y=12z, where z is an integer. Y is a multiple of 12. i.e Y can be 12, 24, 36. And therefore x can be 192, 204 etc.

So if y = 12 then x = 108

Prime factors of 12 = 2^2 *3^1
Prime factors of 108 = 2^2 * 3^3
GCF = 2^2 * 3^1 = 12

Now if y = 24, x = 204

Prime factors of 24 = 2^3 * 3 ^1
Prime factors of 204 = 2^2 * 3^1 * 17
GCF = 2^2 * 3 = 12

So GCF or GCF will be 12 and therefore B alone is sufficient to answer this question. Am I right guys? Unfortunately OA is not provided.
[Reveal] Spoiler: OA

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MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Last edited by Bunuel on 16 Dec 2012, 08:11, edited 2 times in total.
Added the OA
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Re: GCD of x & y [#permalink]  29 Jan 2012, 17:39
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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?

Given: $$x=8y+12$$.

(1) x = 12u, where u is an integer --> $$x=12u$$ --> $$12u=8y+12$$ --> $$3(u-1)=2y$$ --> 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, where z is an integer --> $$y=12z$$ --> $$x=8*12z+12$$ --> $$x=12(8z+1)$$. So, we have $$y=12z$$ and $$x=12(8z+1)$$. Now, as $$z$$ and $$8z+1$$ do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of $$x$$ and $$y$$. Sufficient.

Answer: B.

Hope it's clear.
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Re: If x and y are positive integers such that x=8y+12 [#permalink]  23 Mar 2012, 12:23
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dddmba2012 wrote:
If x an 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

Merging similar topics. Please ask if anything remains unclear.

Also discussed:
if-x-and-y-are-positive-integers-such-that-x-8y-12-what-101196.html
if-x-and-y-are-positive-integers-such-that-x-8y-12-what-100138.html

Hope it helps.
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  26 Aug 2013, 21:51
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[quote="enigma123"]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.

1) x = 12u --> 12u = 8y + 12 --> y = 3(u - 1)/2
Keeping in mind y is a positive integer, u = 3, 5, 7... ---> x = 36, 60, 84 and y = 3, 6, 9..and GCD of x and y is = 3, 6, 3 etc. Since GCD is not constant we cannot determine it.

2) y = 12z ---> x = 8 Ã— 12z + 12 = 12(8z + 1). Now z = 1, 2, 3, 4... ---> y = 12, 24, 36, 48... and x = 12 Ã— 9, 12 Ã— 17, 12 Ã— 25...you can see that GCD is 12 for every pair of x and y.

Hence, 2 answers the question.

Source: http://totalgadha.com/mod/forum/discuss.php?d=130
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  30 Aug 2014, 13:51
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  06 Dec 2014, 07:16
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dmgmat2014 wrote:
Hi Bunuel,

Please can you identify the gap in my understanding?

x= 8y + 12
x = 4(2y+3)

From 1: x = 12 u => x = 4 X 3 X U
This means that (2y+3) must be a multiple of 3. The only way this can happen is if y is a multiple of 3. Lets say y = 3z

x = 4 X 3 X (2z+1)

y = 3 z

z and 2z+1 are co-prime.

So the HCF is 3.

What if z and 4 have some common factors? For example, consider z=2.
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  12 Feb 2012, 22:35
bunuel where enigma123 is wrong in her explanation , i think her way is also correct, by putting values we can easily get to know relevant options, i think by substitiuing varoius values of Y like Y= 12, 24, 36 it becomes little bit lengthy , plz correct me if i am wrong.

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If x and y are positive integers such that x=8y+12 [#permalink]  23 Mar 2012, 12:17
If x an 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
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  05 Jul 2013, 01:44
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  20 Sep 2014, 08:44
Bunuel wrote:
If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?

Given: $$x=8y+12$$.

(1) x = 12u, where u is an integer --> $$x=12u$$ --> $$12u=8y+12$$ --> $$3(u-1)=2y$$ --> 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, where z is an integer --> $$y=12z$$ --> $$x=8*12z+12$$ --> $$x=12(8z+1)$$. So, we have $$y=12z$$ and $$x=12(8z+1)$$. Now, as $$z$$ and $$8z+1$$ do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of $$x$$ and $$y$$. Sufficient.

Answer: B.

Hope it's clear.

Bunuel
Are (kq + 1 , q) always co-primes? where k and q are any positive integers?
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  20 Sep 2014, 12:35
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tushain wrote:
Bunuel wrote:
If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?

Given: $$x=8y+12$$.

(1) x = 12u, where u is an integer --> $$x=12u$$ --> $$12u=8y+12$$ --> $$3(u-1)=2y$$ --> 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, where z is an integer --> $$y=12z$$ --> $$x=8*12z+12$$ --> $$x=12(8z+1)$$. So, we have $$y=12z$$ and $$x=12(8z+1)$$. Now, as $$z$$ and $$8z+1$$ do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of $$x$$ and $$y$$. Sufficient.

Answer: B.

Hope it's clear.

Bunuel
Are (kq + 1 , q) always co-primes? where k and q are any positive integers?

Yes. kq and kq + 1 are consecutive integers, thus they do not share any common factor but 1, thus q and kq + 1 must also be co-prime.
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  06 Dec 2014, 06:53
Hi Bunuel,

Please can you identify the gap in my understanding?

x= 8y + 12
x = 4(2y+3)

From 1: x = 12 u => x = 4 X 3 X U
This means that (2y+3) must be a multiple of 3. The only way this can happen is if y is a multiple of 3. Lets say y = 3z

x = 4 X 3 X (2z+1)

y = 3 z

z and 2z+1 are co-prime.

So the HCF is 3.
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Re: x and y are positive integers such that x=8y+12, what is the [#permalink]  06 Dec 2014, 10:28
Bunuel wrote:
dmgmat2014 wrote:
Hi Bunuel,

Please can you identify the gap in my understanding?

x= 8y + 12
x = 4(2y+3)

From 1: x = 12 u => x = 4 X 3 X U
This means that (2y+3) must be a multiple of 3. The only way this can happen is if y is a multiple of 3. Lets say y = 3z

x = 4 X 3 X (2z+1)

y = 3 z

z and 2z+1 are co-prime.

So the HCF is 3.

What if z and 4 have some common factors? For example, consider z=2.

Thank you. I knew I was missing something
Re: x and y are positive integers such that x=8y+12, what is the   [#permalink] 06 Dec 2014, 10:28
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