Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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. _________________

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.

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.. _________________

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.

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!!

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]

Show Tags

14 Mar 2012, 09:08

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]

Show Tags

26 Jun 2013, 08:14

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.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...