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Re: If a and b are positive integers [#permalink]
02 Sep 2010, 22:41

1

This post received KUDOS

note that a=3b does not mean that 6 will be the gcd. example 6, 18 is 6, but 12, 36 is 12 not suff

for a =2b+6= 2(b+3); take b=6,a=18, gcd =6 ; take b=12,a=30; gcd=6; take b=18,a=42; gcd=6 suff

for an algebraic proof: a=6r, b=6s 2. a=3b means 6r=18s or r=3s; a=6s b=18s; cannot conclude about gcd a=2b+6 means r=2s+1; a=6(2s+1); b=6s no common factors; you can conclude that 6 will be gcd. _________________

Re: If a and b are positive integers [#permalink]
03 Sep 2010, 05:20

10

This post received KUDOS

Expert's post

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This post was BOOKMARKED

gmatbull 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

What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?

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.

Re: If a and b are positive integers [#permalink]
16 Feb 2011, 23:26

Bunuel wrote:

gmatbull 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

What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?

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.

Hope it helps.

Thanks for the rule. I picked A as I knew there is "some" rule for common multiples a an integer with that integer as the difference between the common multiples but was not recollecting it.. _________________

Re: If a and b are positive integers [#permalink]
17 Feb 2011, 17:33

Very good approach to alternative 1.

Bunuel wrote:

gmatbull 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

What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?

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.

Re: If a and b are positive integers [#permalink]
05 Feb 2012, 01:43

Bunuel wrote:

gmatbull 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

What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?

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.

Hope it helps.

But if a and b are both divisible of 6, means that both are even, therefore at least both of them should be divisible by 2.... I am right????? I do not understand why (1) is valid... thanks!!

Re: If a and b are positive integers [#permalink]
05 Feb 2012, 02:02

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Saurajm wrote:

But if a and b are both divisible of 6, means that both are even, therefore at least both of them should be divisible by 2.... I am right????? I do not understand why (1) is valid... thanks!!

Yes, both are divisible by 6, which means that they are divisible by 2 and 3.

Next, 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 \(GCD(a,b)=6\).

Or another way: \(b=6y\) and \(a=6(2y+1)\). \(2y\) and \(2y+1\) are consecutive integers and consecutive integers do not share any common factor 1. As \(2y\) has all factors of \(y\) then \(y\) and \(2y+1\) also do not share any common factor but 1, which means that 6 must GCD of \(a\) and \(b\)

Re: If a and b are positive integers divisible by 6, is 6 the [#permalink]
14 Mar 2012, 08:08

Expert's post

pavanpuneet wrote:

Hi Bunuel, Can you explain the following rule with few examples.

Given: and . Question: is ? 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.

Sure. Both \(a\) and \(b\) are multiples of 6 --> \(a=6x\) and \(b=6y\). Consider two cases:

A. \(x\) and \(y\) do not share any common factor >1, for example \(a=6*2=12\) and \(b=6*3=18\) --> GCD(a,b)=6. As you can see 2 and 3 did not contribute any common factor to the GCD;

B. \(x\) and \(y\) share some common factor >1, for example \(a=6*2=12\) and \(b=6*4=24\) --> GCD(a,b)=12, here 2 and 4 contributes common factor 2 to the GCD.

Re: If a and b are positive integers divisible by 6, is 6 the [#permalink]
26 Jun 2013, 07:14

1

This post received KUDOS

gmatbull 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

What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?

Question: 6 * x =a and 6 * y =b we need to find if 6 is the GCD? YES or NO question. so, basically if we can find a single common factor in x and y, thats it its not a GCD, or if we cant find one then that should also work for us.

(1) a = 2b + 6

6x=12y + 6

x = 2y +1 =>No matter what you do, this will always result in an no common factor.

Thus 6 is the only GCD =>Sufficient

(2) a = 3b

6x = 18y x=3y

Take y=8, and Y=2 =>This is clearly Not sufficient.

Ans: A _________________

PS: Like my approach? Please Help me with some Kudos.

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