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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?

Thank you.

gmatclubot

Re: GCD 2 (Tougher)
[#permalink]
19 Nov 2013, 00:24

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