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

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Intern
Joined: 23 Dec 2014
Posts: 9
If x and y are positive integers such that x = 8y + 12, what is the [#permalink]

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21 Aug 2017, 09:00
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.

Thanks
Manoj Parashar
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Joined: 14 May 2016
Posts: 23
Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]

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07 Apr 2018, 07:19
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.

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?
Intern
Joined: 30 Nov 2017
Posts: 31
If x and y are positive integers such that x = 8y + 12, what is the [#permalink]

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 45455
Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]

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17 May 2018, 04:54
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Expert's post
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] 17 May 2018, 04:54

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