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

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

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17 Sep 2010, 04:23
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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?

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

2) y = 12z, where z is an integer.

well I guess the first questions was quite easy. How about this one? do you still use numbers to solve?
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Re: GCD 2 (Tougher) [#permalink]

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17 Sep 2010, 04:26
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rafi 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.

well I guess the first questions was quite easy. How about this one? do you still use numbers to solve?

OK. Algebraic approach:

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

(1) $$x=12u$$ --> $$12u=8y+12$$ --> $$3(u-1)=2y$$ --> 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) $$y=12z$$ --> $$x=8*12z+12$$ --> $$x=12(8z+1)$$ --> so 12 is a factor both $$x$$ and $$y$$.

Is it GCD of $$x$$ and $$y$$? Why can not it be more than 12, for example 13, 16, 24, ... We see that factors of $$x$$ are 12 and $$8z+1$$: so if $$8z+1$$ has some factor >1 common with $$z$$ then GCD of $$x$$ and $$y$$ will be more than 12 (for example if $$z$$ and $$8z+1$$ are multiples of 5 then $$x$$ would be multiple of $$12*5=60$$ and $$y$$ also would be multiple of $$12*5=60$$, so GCD of $$x$$ and $$y$$ would be more than 12). But $$z$$ and $$8z+1$$ CAN NOT share any common factor >1, as $$8z+1$$ is a multiple of $$z$$ plus 1, so no factor of $$z$$ will divide $$8z+1$$ evenly, which means that GCD of $$x$$ and $$y$$ can not be more than 12. $$GCD(x,y)=12$$. Sufficient.

Answer: B.

Hope it's clear.
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Re: GCD 2 (Tougher) [#permalink]

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17 Sep 2010, 04:30
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...
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Re: GCD 2 (Tougher) [#permalink]

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17 Sep 2010, 04:33
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An easy way to show that z and 8z+1 have no common factors is to use :

$$gcd(a,b) = gcd(a-b,b)$$ when a>b

So gcd(z,8z+1)=gcd(z,1)=1
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Re: GCD 2 (Tougher) [#permalink]

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17 Sep 2010, 04:35
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rafi wrote:
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...

There is a general rule: if $$a$$ and $$b$$ are multiples of $$k$$ and are $$k$$ units apart from each other then $$k$$ is greatest common divisor of $$a$$ and $$b$$.

For example if $$a$$ and $$b$$ are multiples of 7 and $$a=b+7$$ then 7 is GCD of $$a$$ and $$b$$.

So if we apply this rule to (2) we would have: both $$x$$ and $$y$$ are multiple of 12 and are 12 apart each other, so 12 is GCD of $$x$$ and $$y$$.

So in my previous post I just showed the way this general rule is derived.

Hope it helps.
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Re: GCD 2 (Tougher) [#permalink]

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17 Sep 2010, 10:49
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Bunuel wrote:
rafi wrote:
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...

There is a general rule: if $$a$$ and $$b$$ are multiples of $$k$$ and are $$k$$ units apart from each other then $$k$$ is greatest common divisor of $$a$$ and $$b$$.

For example if $$a$$ and $$b$$ are multiples of 7 and $$a=b+7$$ then 7 is GCD of $$a$$ and $$b$$.

So if we apply this rule to (2) we would have: both $$x$$ and $$y$$ are multiple of 12 and are 12 apart each other, so 12 is GCD of $$x$$ and $$y$$.

So in my previous post I just showed the way this general rule is derived.

Hope it helps.

wow Thanx Bunuel!!
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Re: GCD 2 (Tougher) [#permalink]

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19 Sep 2010, 07:26
shrouded1 wrote:
An easy way to show that z and 8z+1 have no common factors is to use :

$$gcd(a,b) = gcd(a-b,b)$$ when a>b

So gcd(z,8z+1)=gcd(z,1)=1

Cool! I'll definitely use that! thanks!
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GMAT Prep - set of questions [#permalink]

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25 Oct 2010, 07:36
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Re: GCD 2 (Tougher) [#permalink]

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25 Oct 2010, 08:55
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rafi: Same logic as that given by Bunuel and shrouded1 above, just worded differently in case you have come across this before: "Two consecutive integers do not have any common factors other than 1"

So 8z and 8z + 1 will not share any factors other than 1 and all factors of z will be factors of 8z too. Therefore, z and 8z + 1 will not have any common factors other than 1.
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Re: GCD 2 (Tougher) [#permalink]

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19 Nov 2013, 01:24
Bunuel wrote:
rafi wrote:
Wow! I have to come up with this conclusion in 2 minutes?
Thanks! I guess it will help me if I'll see something similar...

There is a general rule: if $$a$$ and $$b$$ are multiples of $$k$$ and are $$k$$ units apart from each other then $$k$$ is greatest common divisor of $$a$$ and $$b$$.

For example if $$a$$ and $$b$$ are multiples of 7 and $$a=b+7$$ then 7 is GCD of $$a$$ and $$b$$.

So if we apply this rule to (2) we would have: both $$x$$ and $$y$$ are multiple of 12 and are 12 apart each other, so 12 is GCD of $$x$$ and $$y$$.

So in my previous post I just showed the way this general rule is derived.

Hope it helps.

Bunuel I have a question with statement 2. Kindly clarify:

if $$y= 12z$$ and $$x= 8*12z+12$$ , then $$x - y = 96z + 12 - 12z = 84z + 12$$.

$$84z + 12$$ is not euqal to $$12$$

how is it that you're saying $$x$$ and $$y$$ are 12 units apart from each other?

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

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24 May 2015, 00:15
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]

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26 May 2015, 13:04
@Bunnel :- Can you please tell how are those 12 units apart from each other??
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]

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26 May 2015, 22:48
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Shree9975 wrote:
@Bunnel :- Can you please tell how are those 12 units apart from each other??

Hi Shree9975 and HKHR,

Since both of you have same doubt, I will take it up together.

In this question x = 8y + 12 i.e. x can be represented as some multiple of y + 12. So, if y is divisible by 3, x will also be divisible by 3, similarly for 4 and 6( as 3,4, 6 divide 12). Also, if y is divisible by 12, x will also be divisible by 12.

Consider a situation where y is divisible by 16, will then x be divisible by 16 too? It will not because 12 is not divisible by 16. In fact if y is divisible by any number greater than 12, x will not be divisible by that number, it will always leave a remainder of 12.

So, here x and y are not 12 units apart but x is 12 units apart from a multiple of y. Since st-II tells us that y = 12z i.e. y is divisible by 12, x will always be divisible by 12 as x is 12 units apart from a multiple of y.

Since y = 12z, x = 12(8z + 1).The only thing we need to be careful here is if z and 8z + 1 have a common factor. 8z + 1 can be again interpreted as some multiple of z + 1. So, if any number greater than 1 is a factor of z, it will always leave a remainder of 1 when dividing 8z + 1. Hence, z and 8z + 1 will not have a common factor greater than 1.

Therefore 12 will be the highest number which divides both x and y i.e. their GCD.

Hope it's clear

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

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26 Aug 2015, 13:08
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Thanks Harsh from EgmatQuantExpert,

Most lucid explanation I have come across.
+Many Kudos
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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] 24 Oct 2016, 05:20
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