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 --> \(x=12u\) --> \(12u=8y+12\) --> \(3(u-1)=2y\) --> 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, 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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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) \(x=12u\) --> \(12u=8y+12\) --> \(3(u-1)=2y\) --> 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 cannot 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 \(8z+1\) CANNOT 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 \(y\) cannot be more than 12. \(GCD(x,y)=12\). Sufficient.

Answer: B.

Hope it's clear.
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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...

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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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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I don't understand the quoted solution, can someone please explain it?

Thanks!
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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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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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Bunuel,

You wrote:

(1) x=12u --> 12u=8y+12 --> 3(u-1)=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?

3(u-1)=2y --> the only thing we know from this is that 3 is a factor of y --> 3(u-1) is a multiple of 3, so must be 2y as they are equal. Now, 2y to be multiple of 3 then y must be multiple of 3.
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Bunuel
Are (kq + 1 , q) always co-primes? 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 co-prime.
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@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
Harsh
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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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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(u-1), 2y=3(u-1), 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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Dear Bunuel, Could you please help to explain why are we considering 8z & 8z+1 in this case?
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We established that \(z\) and \(8z+1\) do not share any common factor but 1 based on the fact that \(8z\) and \(8z+1\) do not share any common factor 1.
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I share the same feeling. Doing really bad with DS. How many DS 700+ are expected in exam?

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