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

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

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New post 21 Aug 2017, 08:00
Hello All,

My take on this question:

As given , x = 8y+12 and x and y are positive integers.

Statement 1: x = 12u, u is integer

It means x is multiple of 12. But x is also 8y+12 then it means 8y should also be multiple of x, which in turn means y can take values 3 , 6, 9 ...etc i.e. multiple of 3.
Case 1: say y = 3 then x = 8*3 + 12 = 36 , GCD(x,y) = 3
Case 2: say y = 6 then x = 8*6 +12 = 60, GCD(x,y) = 6

Clearly 1 insufficient.

Statement 2: y = 12u, u is integer

It means y is multiple of 12 , which in turn means x = 8(multiple of 12) + 12 i.e. x is nothing but the next multiple of 12 than y or y and x are consecutive multiples of 12.
So GCD of two consecutive multiples of 12 should be 12 ,hence sufficient. Answer B.

Hope it helps.

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Manoj Parashar
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Re: If x and y are positive integers such that x = 8y + 12, what is the  [#permalink]

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New post 07 Apr 2018, 06:19
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.



Hello Bunuel,

is it possible to take the approach for Statement 1 when dividing by 4 both sides ----> 3u = 2y + 3 can I say that this is insufficient since we have one equation with two variables?
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If x and y are positive integers such that x = 8y + 12, what is the  [#permalink]

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New post Updated on: 17 May 2018, 03:58
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 \(y\). I

Can you please explain this Bunuel?

3(u-1)=2y

So, we know that 3 is a factor of 2y, but how can we conclude that 3 is a factor of y? Is it because 2 and 3 don't have any factor in common, except 1?

Originally posted by Nived on 17 May 2018, 03:48.
Last edited by Nived on 17 May 2018, 03:58, edited 1 time in total.
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Re: If x and y are positive integers such that x = 8y + 12, what is the  [#permalink]

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New post 17 May 2018, 03:54
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Nived 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 \(y\). I

Can you please explain the Bunuel?

3(u-1)=2y

So, we know that 3 is a factor of 2y, but how can we conclude that 3 is a factor of y? Is it because 2 and 3 don't have any factor in common, except 1?


Exactly. 2y/3 = integer (because u-1 is an integer). This to be true, y must be a multiple of 3.
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Re: If x and y are positive integers such that x = 8y + 12, what is the  [#permalink]

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New post 10 Sep 2018, 06:16
VeritasKarishma JeffTargetTestPrep please share your detail approach to this problem. Thanks
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Re: If x and y are positive integers such that x = 8y + 12, what is the  [#permalink]

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New post 11 Sep 2018, 22:45
enigma123 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.



x = 8y + 12

(1) x = 12u, where u is an integer.

If x is a multiple of 12, it means 8y is a multiple of 12. Since 8 already has three 2s, we NEED y to have a 3 but it COULD have 2s and/or other factors too.
So we cannot say what the GCD of x and y is.
Not sufficient.

(2) y = 12z, where z is an integer.
If y is a multiple of 12, x is a multiple of 12 too. So they certainly have 12 common. Let's see what else they could have common.
x is 12 more than a multiple of y so the only common factors they could have are the factors of 12. We already know that they both have 12 in them. So GCD must be 12.
(This concept has been discussed in detail here: https://www.veritasprep.com/blog/2015/0 ... -the-gmat/)
Sufficient.

Answer (B)
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Re: If x and y are positive integers such that x = 8y + 12, what is the  [#permalink]

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New post 11 Nov 2018, 06:57
enigma123 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.

\(\left\{ \begin{gathered}
x,y\,\, \geqslant \,\,1\,\,{\text{ints}} \hfill \\
x - 8y = 12\,\,\,\,\,\left( * \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,GCD\left( {x,y} \right)\)


\(\left( 1 \right)\,\,\,x = 12u\,\,,\,\,\,u\,\,\operatorname{int} \,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,8y = 12\left( {u - 1} \right)\)

\(\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,u = 3\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\,y = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {x,y} \right) = \left( {12 \cdot 3\,,\,3} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 3 \hfill \\
\,{\text{Take}}\,\,u = 5\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\,y = 6\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {x,y} \right) = \left( {12 \cdot 5\,,\,6} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 6\,\, \hfill \\
\end{gathered} \right.\)


\(\left( 2 \right)\,\,\,y = 12z\,\,,\,\,\,z\,\,\operatorname{int} \,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,x = 12 + 8 \cdot 12 \cdot z = 12\left( {8z + 1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,? = 12\)

\(\left( {**} \right)\,\,\,GCD\,\,\left( {z\,,\,8z + 1} \right) = \,\,k \geqslant 1\,\,\,{\text{int}}\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
\,\frac{z}{k} = {\text{in}}{{\text{t}}_{\text{1}}} \hfill \\
\,\frac{{8z + 1}}{k} = {\operatorname{int} _2}\,\,\,\,\, \hfill \\
\end{gathered} \right. \Rightarrow \,\,\,\,\,\,\,\,\frac{1}{k} = {\operatorname{int} _2} - 8\left( {\frac{z}{k}} \right) = {\operatorname{int} _2} - 8 \cdot {\operatorname{int} _1} = \operatorname{int} \,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\, \geqslant \,1\,\,\,{\text{int}}} \,\,\,\,\,k = 1\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If x and y are positive integers such that x = 8y + 12, what is the &nbs [#permalink] 11 Nov 2018, 06:57

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