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If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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29 Jan 2012, 17:25
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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. For me its B and this is how I solved it. Is my solution correct?
Question is asking for GCD of x and y.
GCF or GCD is the product of common prime factors with lowest exponents. for example GF of 12 and 24 is
12 = 2^2 * 3^1 24 = 2^3 * 3^1
GCF = 2^2 * 3^1 = 12
Coming back to the question and considering statement 1
x is a multiple of 12
So if we put different values of x in the our equation GCF will be different. Therefore this statement is INSUFFICIENT.
Considering statement 2
Y=12z, where z is an integer. Y is a multiple of 12. i.e Y can be 12, 24, 36. And therefore x can be 192, 204 etc.
So if y = 12 then x = 108
Prime factors of 12 = 2^2 *3^1 Prime factors of 108 = 2^2 * 3^3 GCF = 2^2 * 3^1 = 12
Now if y = 24, x = 204
Prime factors of 24 = 2^3 * 3 ^1 Prime factors of 204 = 2^2 * 3^1 * 17 GCF = 2^2 * 3 = 12
So GCF or GCF will be 12 and therefore B alone is sufficient to answer this question. Am I right guys? Unfortunately OA is not provided.
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Last edited by Bunuel on 26 Apr 2016, 08:14, edited 3 times in total.
Added the OA



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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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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?Given: \(x=8y+12\). (1) x = 12u, where u is an integer > \(x=12u\) > \(12u=8y+12\) > \(3(u1)=2y\) > 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, where z is an integer > \(y=12z\) > \(x=8*12z+12\) > \(x=12(8z+1)\). So, we have \(y=12z\) and \(x=12(8z+1)\). Now, as \(z\) and \(8z+1\) do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of \(x\) and \(y\). 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 is the [#permalink]
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12 Feb 2012, 22:35
bunuel where enigma123 is wrong in her explanation , i think her way is also correct, by putting values we can easily get to know relevant options, i think by substitiuing varoius values of Y like Y= 12, 24, 36 it becomes little bit lengthy , plz correct me if i am wrong.
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23 Mar 2012, 12:17
If x an 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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[quote="enigma123"]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. 1) x = 12u > 12u = 8y + 12 > y = 3(u  1)/2 Keeping in mind y is a positive integer, u = 3, 5, 7... > x = 36, 60, 84 and y = 3, 6, 9..and GCD of x and y is = 3, 6, 3 etc. Since GCD is not constant we cannot determine it. 2) y = 12z > x = 8 Ã— 12z + 12 = 12(8z + 1). Now z = 1, 2, 3, 4... > y = 12, 24, 36, 48... and x = 12 Ã— 9, 12 Ã— 17, 12 Ã— 25...you can see that GCD is 12 for every pair of x and y. Hence, 2 answers the question. Source: http://totalgadha.com/mod/forum/discuss.php?d=130



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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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20 Sep 2014, 08:44
Bunuel wrote: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?
Given: \(x=8y+12\).
(1) x = 12u, where u is an integer > \(x=12u\) > \(12u=8y+12\) > \(3(u1)=2y\) > 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, where z is an integer > \(y=12z\) > \(x=8*12z+12\) > \(x=12(8z+1)\). So, we have \(y=12z\) and \(x=12(8z+1)\). Now, as \(z\) and \(8z+1\) do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of \(x\) and \(y\). Sufficient.
Answer: B.
Hope it's clear. BunuelAre (kq + 1 , q) always coprimes? where k and q are any positive integers?



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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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20 Sep 2014, 12:35
tushain wrote: Bunuel wrote: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?
Given: \(x=8y+12\).
(1) x = 12u, where u is an integer > \(x=12u\) > \(12u=8y+12\) > \(3(u1)=2y\) > 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, where z is an integer > \(y=12z\) > \(x=8*12z+12\) > \(x=12(8z+1)\). So, we have \(y=12z\) and \(x=12(8z+1)\). Now, as \(z\) and \(8z+1\) do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of \(x\) and \(y\). Sufficient.
Answer: B.
Hope it's clear. BunuelAre (kq + 1 , q) always coprimes? where k and q are any positive integers? Yes. kq and kq + 1 are consecutive integers, thus they do not share any common factor but 1, thus q and kq + 1 must also be coprime.
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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06 Dec 2014, 06:53
Hi Bunuel,
Please can you identify the gap in my understanding?
x= 8y + 12 x = 4(2y+3)
From 1: x = 12 u => x = 4 X 3 X U This means that (2y+3) must be a multiple of 3. The only way this can happen is if y is a multiple of 3. Lets say y = 3z
x = 4 X 3 X (2z+1)
y = 3 z
z and 2z+1 are coprime.
So the HCF is 3.



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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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06 Dec 2014, 10:28
Bunuel wrote: dmgmat2014 wrote: Hi Bunuel,
Please can you identify the gap in my understanding?
x= 8y + 12 x = 4(2y+3)
From 1: x = 12 u => x = 4 X 3 X U This means that (2y+3) must be a multiple of 3. The only way this can happen is if y is a multiple of 3. Lets say y = 3z
x = 4 X 3 X (2z+1)
y = 3 z
z and 2z+1 are coprime.
So the HCF is 3. What if z and 4 have some common factors? For example, consider z=2. Thank you. I knew I was missing something



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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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07 Nov 2015, 06:09
enigma123 wrote: 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.
x=8y+12 = 4(2y+3) i.e. x is a Multiple of 4 Statement 1: X=12u i.e. x is a multiple of 12 i.e. y must be a multiple of 3 but since y may be an even multiple of 3 or an odd multiple of 3 so GCD will have different values. Hence, NOT SUFFICIENT Statement 2: Y=12z i.e. y must be a multiple of 3 as well 4 for such value of y, x must be a multiple of 12 e.g. @y=12, x = 4*27, GCD = 12 @y=24, x = 4*51, GCD = 12 but since y is an even multiple of 3 so GCD will have constant value. Hence, SUFFICIENT Answer: Option B
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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10 Nov 2015, 10:15
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. 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. There are 2 variables (x,y), one equation (x=8y+12), and 2 more equations are given from the 2 conditions; there is high chance (D) will be our answer. From condition 1, 12u=8y+12, 8y=12(u1), 2y=3(u1), from x=4(2y+3), this has to be a multiple of y=3, but x is a multiple of 4, so we cannot decide the GCD; this is insufficient. From condition 2, x=8(12z)+12=12(8z+1), z cannot equal 8z+1, the GCD(x,y)=12, so this is sufficient, and the answer becomes (B). For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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16 Mar 2016, 08:57
Nice Question .. It fooled Me i choose E Forgot to check the cases for z and 8z+1 as they are coprimes. Hence Answer is B
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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27 Mar 2017, 21:36
Bunuel wrote: tushain wrote: Bunuel wrote: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?
Given: \(x=8y+12\).
(1) x = 12u, where u is an integer > \(x=12u\) > \(12u=8y+12\) > \(3(u1)=2y\) > 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, where z is an integer > \(y=12z\) > \(x=8*12z+12\) > \(x=12(8z+1)\). So, we have \(y=12z\) and \(x=12(8z+1)\). Now, as \(z\) and \(8z+1\) do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of \(x\) and \(y\). Sufficient.
Answer: B.
Hope it's clear. BunuelAre (kq + 1 , q) always coprimes? where k and q are any positive integers? Yes. kq and kq + 1 are consecutive integers, thus they do not share any common factor but 1, thus q and kq + 1 must also be coprime. Dear Bunuel, Could you please help to explain why are we considering 8z & 8z+1 in this case?
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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05 Aug 2017, 18:18
Bunuel wrote: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?
Given: \(x=8y+12\).
(1) x = 12u, where u is an integer > \(x=12u\) > \(12u=8y+12\) > \(3(u1)=2y\) > 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, where z is an integer > \(y=12z\) > \(x=8*12z+12\) > \(x=12(8z+1)\). So, we have \(y=12z\) and \(x=12(8z+1)\). Now, as \(z\) and \(8z+1\) do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of \(x\) and \(y\). Sufficient.
Answer: B.
Hope it's clear. Bunuel You blow my mind! Incredible!
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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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08 Aug 2017, 03:17
1) x=12u u=1,2,3,4...... Than x=12, 36, 48,...... Y= x12/8=0,3,4..... different GCDNot sufficient
2) Y=12z x=8*12z+12=12(8z+1)a z=1, 2, 3, 4..... Than y=12*1, 12*3, 12*4,.... x= 12*9, 12*17, 12*25, 12*33(do not calculate just plugin in a Common between x&y is 12 hence sufficient



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Re: If x and y are positive integers such that x = 8y + 12, what is the [#permalink]
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09 Aug 2017, 19:50
Bunuel wrote: tushain wrote: Bunuel wrote: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?
Given: \(x=8y+12\).
(1) x = 12u, where u is an integer > \(x=12u\) > \(12u=8y+12\) > \(3(u1)=2y\) > 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, where z is an integer > \(y=12z\) > \(x=8*12z+12\) > \(x=12(8z+1)\). So, we have \(y=12z\) and \(x=12(8z+1)\). Now, as \(z\) and \(8z+1\) do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of \(x\) and \(y\). Sufficient.
Answer: B.
Hope it's clear. BunuelAre (kq + 1 , q) always coprimes? where k and q are any positive integers? Yes. kq and kq + 1 are consecutive integers, thus they do not share any common factor but 1, thus q and kq + 1 must also be coprime. This question is beyond esoteric. Not saying, I don't easily understand what coprimes are and how you set up that formula. However, I didn't know KQ+1 and Q are coprimes




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