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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.

Re: If x and y are positive integers such that x=8y+12 [#permalink]
23 Mar 2012, 12:23

1

This post received KUDOS

Expert's post

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.

Re: x and y are positive integers such that x=8y+12, what is the [#permalink]
26 Aug 2013, 21:51

1

This post received KUDOS

[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.

Re: x and y are positive integers such that x=8y+12, what is the [#permalink]
30 Aug 2014, 13:51

1

This post received KUDOS

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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

1

This post received KUDOS

Expert's post

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. _________________

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.

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 _________________

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?

Re: x and y are positive integers such that x=8y+12, what is the [#permalink]
20 Sep 2014, 12:35

Expert's post

1

This post was BOOKMARKED

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. _________________

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

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

gmatclubot

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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