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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?

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.

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.

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.

Re: GCD 2 (Tougher) [#permalink]
25 Oct 2010, 07:55

Expert's post

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. _________________

Re: GCD 2 (Tougher) [#permalink]
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?

Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
23 May 2015, 23:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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 _________________

gmatclubot

Re: If x and y are positive integers such that x = 8y + 12, what
[#permalink]
26 May 2015, 21:48

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