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If a and b are positive integers divisible by 6, is 6 the

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If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

(1) a = 2b + 6
(2) a = 3b

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Originally posted by enigma123 on 22 Jan 2012, 18:05.
Last edited by Bunuel on 23 May 2013, 05:12, edited 2 times in total.
OA added
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Re: GCD of A & B  [#permalink]

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New post 22 Jan 2012, 18:18
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enigma123 wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b

As the OA is not provided, for me its clear C i.e. both statement together are sufficient to say that "Yes" 6 is the GCD of a & b. Can someone please confirm?


If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

Given: \(a=6x\) and \(b=6y\). Question: is \(GCD(a,b)=6\)? Now, if \(x\) and \(y\) share any common factor >1then \(GCD(a,b)\) will be more than 6 if not then \(GCD(a,b)\) will be 6.

(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.

(2) \(a=3b\) --> clearly insufficient.

Answer: A.

There is also 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 (1) \(a=2b+6\) --> both \(a\) and \(2b\) are multiples of 6 and are 6 apart, so GCD of \(a\) and \(2b\) is 6, hence GCD of \(a\) and \(b\) is also 6. Sufficient.

Also discussed here: if-a-and-b-are-positive-integers-divisible-by-6-is-6-the-100324.html
Similar questions:
what-is-the-greatest-common-factor-of-x-and-y-1-x-and-y-are-109273.html
if-x-and-y-are-positive-integers-such-that-x-8y-12-what-101196.html

Hope it helps.
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 23 Jan 2012, 19:15
Hi Bunuel, could you kindly elaborate on the following statement:

"x=2y+1 --> x and y do not share any factor >1, as if they were we would be able to factor out if from 2y+1. Sufficient."

I'm obviously missing some fundamental insight, but I don't understand why not being able to factor something out of the 2y+1 means that 6 is the GCD.

Thanks!
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 23 Jan 2012, 19:38
1
svalan wrote:
Hi Bunuel, could you kindly elaborate on the following statement:

"x=2y+1 --> x and y do not share any factor >1, as if they were we would be able to factor out if from 2y+1. Sufficient."

I'm obviously missing some fundamental insight, but I don't understand why not being able to factor something out of the 2y+1 means that 6 is the GCD.

Thanks!


We have that \(a=6x\) and \(b=6y\).

Consider two cases:
1. \(x\) and \(y\) share some common factor >1: for example \(x=2\) and \(y=4\) then \(a=12\) and \(b=24\) --> \(GCD(a,b)=12>6\);
2. \(x\) and \(y\) DO NOT share any common factor >1: for example \(x=5\) and \(y=2\) then \(a=30\) and \(b=12\) --> \(GCD(a,b)=6\).

From (1) we have that --> \(x=2y+1\) --> \(x\) is one more than multiple of \(y\). For example: \(x=3\) and \(y=1\) OR \(x=5\) and \(y=2\) OR \(x=7\) and \(y=3\) ... as you can see in all these cases x and y do not share any common factor more than 1. Now, as we concluded above if \(x\) and \(y\) DO NOT share any common factor >1, then \(GCD(a,b)=6\).

Hope it's clear.
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 23 Jan 2012, 19:47
Crystal clear, thanks so much!
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 25 Jan 2012, 10:26
Perfect explanation
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 19 Oct 2013, 08:54
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Bunuel wrote:
svalan wrote:
Hi Bunuel, could you kindly elaborate on the following statement:

"x=2y+1 --> x and y do not share any factor >1, as if they were we would be able to factor out if from 2y+1. Sufficient."

I'm obviously missing some fundamental insight, but I don't understand why not being able to factor something out of the 2y+1 means that 6 is the GCD.

Thanks!


We have that \(a=6x\) and \(b=6y\).

Consider two cases:
1. \(x\) and \(y\) share some common factor >1: for example \(x=2\) and \(y=4\) then \(a=12\) and \(b=24\) --> \(GCD(a,b)=12>6\);
2. \(x\) and \(y\) DO NOT share any common factor >1: for example \(x=5\) and \(y=2\) then \(a=30\) and \(b=12\) --> \(GCD(a,b)=6\).

From (1) we have that --> \(x=2y+1\) --> \(x\) is one more than multiple of \(y\). For example: \(x=3\) and \(y=1\) OR \(x=5\) and \(y=2\) OR \(x=7\) and \(y=3\) ... as you can see in all these cases x and y do not share any common factor more than 1. Now, as we concluded above if \(x\) and \(y\) DO NOT share any common factor >1, then \(GCD(a,b)=6\).

Hope it's clear.


Here's my humble opinion of Bunuel's Explain:

(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.

\(6x=2*6y+6\)
\(6x=12y+6\)
\(6x=6(2y+1)\)
\(x=2y+1\) Here, both sides of the equation are divided by 6, which in this case act as the GCF.

Now talking about whether x and y will have any other common factors when in \(x=2y+1\) form, keep a simple rule in mind:

Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now extrapolating this rule to consecutive multiples: let's say that of 7
7(1)=7, 7(2)=14, 7(3)=21, 7(4)=28, 7(5)=35
Prime factorization of \(7=7^1\)
Prime factorization of \(14+1=15=3^1*5^1 and 14-1=13=1*13^1\)
Prime factorization of \(21+1=22=2^1*11^1 and 21-1=20=2^2*5^1\)
Prime factorization of \(28+1=29=29^1 and 28-1=27=3^3\)
Prime factorization of \(35+1=36=2^2*3^2 and 35-1=34=2^1*17^1\)

So we can say When 1 is added to or subtracted from any Consecutive multiple of N i.e. \(n and (n+1) or (n-1) and n\), the result is a co-prime.


Comments please!
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 14 Jan 2014, 22:08
Bunuel,

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.

I understand this using numbers,but can you please explain it using some theory as you always do?

I mean like 2 and 3 are co prime, i understand that very clearly but the same level of clarity is not here.

Also, how does the statement in bold exactly is derived.

Thanks in advance
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 16 Jan 2014, 07:44
Hi Bunuel,

Can you please explain

x and y do not share any factor >1, as if they were we would be able to factor out if from 2y+1. Sufficient.


so if we have x=2(2y+1) then x does have a factor of 2,but how do we know about the factors of y.
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 16 Jan 2014, 08:22
shankar245 wrote:
Hi Bunuel,

Can you please explain

x and y do not share any factor >1, as if they were we would be able to factor out if from 2y+1. Sufficient.


so if we have x=2(2y+1) then x does have a factor of 2,but how do we know about the factors of y.


x=2y+1 means that x and 2y are co-prime (x and 2y do not share any common factor but 1), which, on the other hand, means that x and y are co-prime. How else? If x and 2y do not share any common factor greater than 1, how can x and y share any common factor greater than 1?
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 01 May 2016, 11:55
If x and 2y do not share any common factor greater than 1, how can x and y share any common factor greater than 1?
Please Bunuel clarify this comment more. what does it mean.
another question is that how frequently DS questions of divisibility come in GMAT test ?
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 02 May 2016, 04:05
hatemnag wrote:
If x and 2y do not share any common factor greater than 1, how can x and y share any common factor greater than 1?
Please Bunuel clarify this comment more. what does it mean.
another question is that how frequently DS questions of divisibility come in GMAT test ?


Let me ask you: if x and y shared any common factor but 1, would x and 2y be co-prime?
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If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 23 Oct 2016, 02:22
Thanks for the explanation.

Still I do not understand why statement 2 is not sufficient. My reasoning is this one:
- for 6 to be GCD(a,b) then "a" and "b" must be multiple of 6 but "a" must also be equal to b+6 right?
- The only way to have a=3b and a=b+6 is for b=3 and a=9. However as b/6 must result in an integer we know that "a" cannot be equal to 3b, hence it is enough to say that 6 is NOT the gcd of (a,b). Therefore statement 2 is sufficient to answer the question.

Can you please tell me where my reasoning is flawed?
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Re: If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 23 Oct 2016, 09:09
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camilles wrote:
Thanks for the explanation.

Still I do not understand why statement 2 is not sufficient. My reasoning is this one:
- for 6 to be GCD(a,b) then "a" and "b" must be multiple of 6 but "a" must also be equal to b+6 right?
- The only way to have a=3b and a=b+6 is for b=3 and a=9. However as b/6 must result in an integer we know that "a" cannot be equal to 3b, hence it is enough to say that 6 is NOT the gcd of (a,b). Therefore statement 2 is sufficient to answer the question.

Can you please tell me where my reasoning is flawed?


Take b = 12, you will get a = 36.

Now try to calculate the HCF of a and b, you will get your answer, why B is insufficient. :)
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If a and b are positive integers divisible by 6, is 6 the  [#permalink]

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New post 27 Mar 2017, 23:13
Bunuel wrote:
enigma123 wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b

As the OA is not provided, for me its clear C i.e. both statement together are sufficient to say that "Yes" 6 is the GCD of a & b. Can someone please confirm?


If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

Given: \(a=6x\) and \(b=6y\). Question: is \(GCD(a,b)=6\)? Now, if \(x\) and \(y\) share any common factor >1then \(GCD(a,b)\) will be more than 6 if not then \(GCD(a,b)\) will be 6.

(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.

(2) \(a=3b\) --> clearly insufficient.

Answer: A.

There is also 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 (1) \(a=2b+6\) --> both \(a\) and \(2b\) are multiples of 6 and are 6 apart, so GCD of \(a\) and \(2b\) is 6, hence GCD of \(a\) and \(b\) is also 6. Sufficient.


Dear Bunuel, What is the reasoning behind the general rule? I have a hard time to get the understanding for this concept.

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.
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