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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] New post 17 Sep 2010, 03:23
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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] New post 17 Sep 2010, 03:26
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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] New post 17 Sep 2010, 03:30
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] New post 17 Sep 2010, 03:33
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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] New post 17 Sep 2010, 03:35
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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] New post 17 Sep 2010, 09:49
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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] New post 19 Sep 2010, 06:26
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] New post 25 Oct 2010, 06:36
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Re: GCD 2 (Tougher) [#permalink] New post 25 Oct 2010, 07:55
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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] New post 19 Nov 2013, 00:24
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] New post 23 May 2015, 23:15
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink] New post 26 May 2015, 12:04
@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] New post 26 May 2015, 21:48
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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
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Re: If x and y are positive integers such that x = 8y + 12, what   [#permalink] 26 May 2015, 21:48
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