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Divisibility & Primes - GCF (Tricky)

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Divisibility & Primes - GCF (Tricky)

by Jayanth2689 » Fri Jun 10, 2011 6:42 am
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

OA is A

Official Explanation for this -

[spoiler]MGMAT says that for a and b to always have 6 as the GCF, they have to be mutually prime.

Hence, Statement 1 is sufficient to answer. This can be tested by plugging in numbers also.

My question is, is it right for to assume that any question which asks to check if a number is the GCF of two numbers can be solved using the mutually prime property?

Or is this dependent on the statements given? in this case statement A. Also, please do share other useful divisibility and primes properties that can be applied for DS questions.[/spoiler]
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by phanideepak » Fri Jun 10, 2011 7:06 am
We know that both a and b are divisible by 6 so lets assume that a = 6x and b = 6y

Now all we have to do is prove that x and y have no common factors except 1

1 : a = 2b +6

6x = 12y +6

so x = 2y + 1

Intuitively u can see that for ANY value of y, x & Y can never have the same factors except 1. You can check by substitution too

for example y = 1 so x = 3 no common factors

y = 2 x = 5 again no common factors

y = 7 x = 15 again no common factors

So A is sufficient

B : a = 3b

lets take a,b = 6,18

in this case the GCF is 6

But if we take a,b = 12,36

here the GCF is 12

So B alone is not sufficient.

so IMO the answer is A
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by manpsingh87 » Fri Jun 10, 2011 7:07 am
Jayanth2689 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

OA is A

Official Explanation for this -

[spoiler]MGMAT says that for a and b to always have 6 as the GCF, they have to be mutually prime.

Hence, Statement 1 is sufficient to answer. This can be tested by plugging in numbers also.

My question is, is it right for to assume that any question which asks to check if a number is the GCF of two numbers can be solved using the mutually prime property?

Or is this dependent on the statements given? in this case statement A. Also, please do share other useful divisibility and primes properties that can be applied for DS questions.[/spoiler]
a=6k; b=6m;
1) a=2b+6;
b=6m;
a=12m+6;
=6(2m+1);
now for different values of m , 6m and 6(2m+1) will have 6 as a GCD, hence 1 alone is sufficient to answer the question.

2)
a=3b;
=3*6m;
=18m;
b=6m,
now for m=1 GCD is 6, and for m=2 GCD is 12 hence 2 alone is not sufficient to answer the question..

therefore answer should be A
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by phanideepak » Fri Jun 10, 2011 7:09 am
Jayanth may be by meaning mutually prime numbers he means that the numbers are co-primes

Co-primes are the numbers which have no common factors except 1.

here we know that a = 6x and b= 6y so if we prove that x and y are co-primes we have the solution and I have done the same in my solution above. I forgot the term co-primes and thank you for helping me remember that :)
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by Jayanth2689 » Fri Jun 10, 2011 7:15 am
phanideepak wrote:Jayanth may be by meaning mutually prime numbers he means that the numbers are co-primes

Co-primes are the numbers which have no common factors except 1.

here we know that a = 6x and b= 6y so if we prove that x and y are co-primes we have the solution and I have done the same in my solution above. I forgot the term co-primes and thank you for helping me remember that :)
yes phanideepak, i was intending the same!
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by cans » Fri Jun 10, 2011 8:18 am
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
b)a=3b
b is the GCD. if b=6, GCD=6 and if b=12;GCD=12
Insufficient
a)a=2b+6
b=6k (where k is an integer)
a=6(2k+1)
k and (2k+1) have G.C.D of 1
Thus a,b GCD = 6
IMO A
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by Jayanth2689 » Fri Jun 10, 2011 8:26 am
cans 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
b)a=3b
b is the GCD. if b=6, GCD=6 and if b=12;GCD=12
Insufficient
a)a=2b+6
b=6k (where k is an integer)
a=6(2k+1)
k and (2k+1) have G.C.D of 1
Thus a,b GCD = 6
IMO A
ah..i did not look at this way!
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by Jayanth2689 » Sun Feb 12, 2012 5:55 am
Hi experts! just had a quick question regarding the problem in this thread (posted by me 6 months back)

If 6 IS the GCF of a and b, how can a and b be mutually prime?

Don't mutually prime numbers have just 1 as a common factor? In this case they also have 2 and 3 as common factors too!

Please do correct me if i'm wrong!
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by Mike@Magoosh » Sun Feb 12, 2012 9:39 pm
This a reply to Jayanth2689's question

First of all, I want to say: great work in this thread! I agree with the answer of A that all participants have gotten and have explained very well.

In the original post, you said:
MGMAT says that for a and b to always have 6 as the GCF, they have to be mutually prime.

I think you might have been misreading/misquoting something from the MGMAT solution there.

If a and b have a GCF of 6, then a and b are most definitely not mutually prime.

BUT, if a and b have a GCF of 6, and if 6 the greatest common divisor of a and b, then that means if we write a = 6x and b = 6y, then we know that x and y must be mutually prime.

Mutually prime does play an important role in the problem, but it applies to x & y, not to a & b.

Does that make sense?

Mike :)
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by [email protected] » Wed Feb 15, 2012 11:14 pm
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

OA is A



Basically the first statement says a clear no for whatever the numbers are as they are a set of co-primes...

Statement 2, the value changes from something to something...
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by lunarpower » Fri Feb 17, 2012 6:22 am
i received a private message regarding this question.

mike --
Mike@Magoosh wrote:BUT, if a and b have a GCF of 6, and if 6 the greatest common divisor of a and b, then that means if we write a = 6x and b = 6y, then we know that x and y must be mutually prime.

Mutually prime does play an important role in the problem, but it applies to x & y, not to a & b.

Does that make sense?

Mike :)
this is correct -- and it's also what's in the MGMAT answer key. therefore, if the original poster somehow inferred that a and b themselves are relatively prime, then the original poster must have misread the explanation.
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by Jayanth2689 » Fri Feb 17, 2012 6:58 am
Mike@Magoosh wrote:This a reply to Jayanth2689's question

First of all, I want to say: great work in this thread! I agree with the answer of A that all participants have gotten and have explained very well.

In the original post, you said:
MGMAT says that for a and b to always have 6 as the GCF, they have to be mutually prime.

I think you might have been misreading/misquoting something from the MGMAT solution there.

If a and b have a GCF of 6, then a and b are most definitely not mutually prime.

BUT, if a and b have a GCF of 6, and if 6 the greatest common divisor of a and b, then that means if we write a = 6x and b = 6y, then we know that x and y must be mutually prime.

Mutually prime does play an important role in the problem, but it applies to x & y, not to a & b.

Does that make sense?

Mike :)
Thanks Mike! Yes it does!!
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by Jayanth2689 » Fri Feb 17, 2012 6:59 am
lunarpower wrote:i received a private message regarding this question.

mike --
Mike@Magoosh wrote:BUT, if a and b have a GCF of 6, and if 6 the greatest common divisor of a and b, then that means if we write a = 6x and b = 6y, then we know that x and y must be mutually prime.

Mutually prime does play an important role in the problem, but it applies to x & y, not to a & b.

Does that make sense?

Mike :)
this is correct -- and it's also what's in the MGMAT answer key. therefore, if the original poster somehow inferred that a and b themselves are relatively prime, then the original poster must have misread the explanation.
Thank you Ron! Yes it seems i had misunderstood/misread the MGMAT explanation! my bad! thanks again!
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by lunarpower » Fri Feb 17, 2012 7:01 am
you're welcome.
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