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

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

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New post 05 Mar 2018, 01:03
Mudit27021988 wrote:
victory47 wrote:
I fail

not easy at all

I want to follow this posting.


I share the same feeling. Doing really bad with DS. How many DS 700+ are expected in exam?

Posted from my mobile device


There is no fixed number of DS questions on the test. It varies from approximately 13 to 17.

3. Strategies and Tactics for DS Section



Hope it helps.
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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, 04:58
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, 04:48.
Last edited by Nived on 17 May 2018, 04:58, 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, 04:54
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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 11 Sep 2018, 23:45
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, 07:57
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  [#permalink]

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Re: If x and y are positive integers such that x = 8y +12 . What   [#permalink] 17 Oct 2019, 07:36

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