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If x and y are positive integers such that x = 8y + 12, what [#permalink]
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31 Aug 2010, 01:22
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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
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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31 Aug 2010, 02:14
Stat 1: x= 12u returns x= 12u and y=3/2(u1) GCD of x and y varies for u=0 and u is +ve
Stat 2: y=12z returns x=12(8z+1),y=12z GCD for any inter value of z is 12.
Hence statement 2 alone is sufficient.
answer:B



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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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31 Aug 2010, 04:52
metallicafan 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 Given: \(x=8y+12\). (1) \(x=12u\) > \(12u=8y+12\) > \(3(u1)=2y\) > 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: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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31 Aug 2010, 13:14
Wow, is that a 700 question?
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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21 Sep 2010, 12:32
kalrac wrote: Stat 1: x= 12u returns x= 12u and y=3/2(u1) GCD of x and y varies for u=0 and u is +ve
Stat 2: y=12z returns x=12(8z+1),y=12z GCD for any inter value of z is 12.
Hence statement 2 alone is sufficient.
answer:B I don't understand the quoted solution, can someone please explain it? Thanks!
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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21 Sep 2010, 12:39
metallicafan wrote: kalrac wrote: Stat 1: x= 12u returns x= 12u and y=3/2(u1) GCD of x and y varies for u=0 and u is +ve
Stat 2: y=12z returns x=12(8z+1),y=12z GCD for any inter value of z is 12.
Hence statement 2 alone is sufficient.
answer:B I don't understand the quoted solution, can someone please explain it? Thanks! I think the quoted solution refers to the following 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 \(y\).
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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16 Oct 2010, 08:40
great question...



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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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16 Feb 2011, 14:23
Bunuel,
You wrote:
(1) x=12u > 12u=8y+12 > 3(u1)=2y > the only thing we know from this is that 3 is a multiple of y.
How do we know that 3 is a muliple of y? I mean, I worked it out by plugging in values for u and found that it is true, but is there some property of the formula that gives it away?



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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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16 Feb 2011, 14:34



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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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01 Mar 2013, 04:39
I fail not easy at all I want to follow this posting.
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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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25 Mar 2014, 10:38
x = 8y + 12 simplifies to x=4(2y+3)
Statement 1: substituting x=12u 12u=4(2y+3) 3u=2y3 3(u1)=2y
So u should be greater than 1 and should be odd (because u1 is a multiple of 2) Checking for values
u x y GCD 3 36 3 3 5 60 6 6
So statement 1 not sufficient
Statement 2: substituting y=12z
x=4(12z+3) x=12(4z+1)
Checking for values
z x y GCD 1 60 12 12 2 108 24 12 3 154 36 12 we can check for more values but usually we would see any variations withing 1st 34 values.
Answer is B.



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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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03 Mar 2018, 00:36
victory47 wrote: I fail
not easy at all
I want to follow this posting. I share the same feeling. Doing really bad with DS. How many DS 700+ are expected in exam? Posted from my mobile device



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Re: If x and y are positive integers such that x = 8y + 12, what [#permalink]
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05 Mar 2018, 01:03




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