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:

Can someone please help and let me know how to approach this question?

My approach is:

Considering statement 1: Prime factors of 12 are: 2,2 & 3. As the GCF of z and 12 is 3, z should be a multiple of 3. Now if its a multiple of 3 it could be 3, 6, 12, 18..... which will be give Yes and No answers to the question and therefore insufficient. But in the book it says it sufficient. Where I am going wrong?

Considering statement 2: Prime factors of 15 are 3 and 5. ...Again I struggle to complete. Can someone please help?

1. The GCF of z and 12 is 3. 2. The GCF of z and 15 is 15.

st-1 is tricky - if the GCF is given as 3, then value of z cannot be 6,12 or any other multiple of 2, because the GCF then would not be 3. So value of Z could be 3, 9, 15, 21 - in all cases the GCF of z and 12 is 3. So Z as per st1 is not divisible by 6.

st 2 - obviously can have values of Z as 15, 30, 45 etc. So Z divisible by 6 may be true or may not be true.

Agree and many thanks. On a separate note: If the LCM of a & 12 is 36 what could be the possible values of a?

How to approach such questions? I can only think upto

1. The value cannot be more than 36. 2. The factors of 36 are 2,2,3,3 3. The factors of 12 are 2,2,3.

I struggle when I see a variable like a,x or n. Can you please let me know how can I improve my thought process? And again how to approach this problem?
_________________

Agree and many thanks. On a separate note: If the LCM of a & 12 is 36 what could be the possible values of a?

How to approach such questions? I can only think upto

1. The value cannot be more than 36. 2. The factors of 36 are 2,2,3,3 3. The factors of 12 are 2,2,3.

I struggle when I see a variable like a,x or n. Can you please let me know how can I improve my thought process? And again how to approach this problem?

I recommend you go through MGMAT number properties guide. It really provides a clear picture about how to find HCF and LCM.
_________________

The LCM of 12 and a contains two 2's. Since the LCM contains each prime factor to the power it appears the MOST, we know that a cannot contain more than two 2's.

LCM of 12 and a contain two 3's. But 12 only contains one 3. The 3^2 factor in the LCM must have come from prime factorization of a. Thus we know that a contains exactly two 3's.

Since a must contain exactly two 3's nd can contain no 2's, one 2 or two 2's a could be

3*3=9 3*3*2=18 3*3*2**2=36.

Thus 9,18 and 36 are three values.

I am struggling to understand the concept.
_________________

So the common factors are 2 & 3. How come 12? Any guidance on GCD will be much appreciated. I say this with BIG please.

12= 2^2*3 24= 2^3*3

GCF(12, 24)= Product of minimum power of all common prime factors.

Locate the common prime factors; 2 and 3; Let's check the minimum power of 2; In 12: 2 has a power of 2. In 24; 2 has a power of 3. Here; 2<3 Thus; GCF will have 2^2(The minimum of the two powers)

Now; In 12; 3 has a power of 1. In 24; 3 has a power of 1. Thus, minimum power of 3 is 1; GCF will have 3^1

GCF=2^2*3^1=12

So; what's the GCF of 630 and 240.

630=3^2*5*2*7 240=2^4*5*3

Locate common prime factors; 2, 3 and 5. Locate minimum powers of 2, 3 and 5 in both of these.

630 has 2 3's i.e. 3^2 240 has 1 3 i.e. 3^1 Thus, we consider: 3^1 for GCF

630 has 1 2 i.e. 2^1 240 has 4 2's i.e. 2^4 Thus, we consider 2^1 for GCF

630 has 1 5 i.e. 5^1 240 has 1 5 i.e. 5^1 Thus, we consider 5^1 for GCF

So, if we are given that GCF of z and 12 is 3, what do we know about z. 12=2^2*3

We know that z has at least one "3" in its factor AND z has no factor of 2 because even if there is one factor of 2 present in z, the GCF becomes 2^1*3^1=6, invalidating the statement; you see the point ***********************************************
_________________

The LCM of 12 and a contains two 2's. Since the LCM contains each prime factor to the power it appears the MOST, we know that a cannot contain more than two 2's.

LCM of 12 and a contain two 3's. But 12 only contains one 3. The 3^2 factor in the LCM must have come from prime factorization of a. Thus we know that a contains exactly two 3's.

Since a must contain exactly two 3's nd can contain no 2's, one 2 or two 2's a could be

3*3=9 3*3*2=18 3*3*2**2=36.

Thus 9,18 and 36 are three values.

I am struggling to understand the concept.

For LCM you will have to consider all prime factors and maximum powers.

LCM(a, 12)=36

12=2^2*3 a=? 36=2^2*3^3 --------------

What does this tell about a?

LCM always has maximum power of the factor; Thus if LCM is 36 and its factors are 2^2*3^2. It means that a only has a maximum of two distinct prime factors 2 and 3 and the maximum powers of those factors are 2 and 2 respectively.

Now, let's see what 12 tells us; 12=2^2*3 Means; a can have 2^0, 2^1 or 2^2 as its factor because the minimum criteria for 36 to have at least 2^2 has already been taken care by 12. Thus, it really doesn't matter whether a contains 2^2 or not. a may contain 2^0, 2^1 or 2^2. Note a can't contain 2^3 because in 36, maximum power of 2 is 2. Thus, any of the numbers can't have more than 2 2's.

Likewise; let's check for 3. 12 has 1 3. But 36 has two 3's i.e. 3^2 Thus, a must contain 3^2; because 36 is LCM of a and 12. As 12 doesn't have 2 factors of 3. It's become necessary for a to have 2 3's. Thus, a has 3^2. Also, note that a can't contain more than 2 3's because 36 has maximum of 2 3's. Also, a can't contain any other prime factor as 36 has only two distinct factors; 3 and 2.

Now, how many values of a are possible; 2^0*3^2=9 2^1*3^2=18 2^2*3^2=36 ****************************************
_________________

1.Sufficient GCF of z and 12 is 3. that tells us that Z doesn't have any 3 , but has a 2.

for a number to be divisible by 6 , it needs to have both 2 and 3 as factors. in the above as 3 is ruled out, we can clearly say that the number is not divisible by 6.

2. Not sufficient GCF of z and 15 is 15 . That tells us that z has 3 and 5 as factors.

But we dont whether there is 2 in it or not. If z has 2 as a factor it is divisible by 6 or else not.

Realistic GMAT question would mention that z is a positive integer.

(1) The greatest common factor of z and 12 is 3 --> if z were divisible by 6 (for example 6, 12, 18, ...) then the GCF of z and 12 (which is also divisible by 6) would have been more than 3 (6 or 12) and since the GCF is 3 then z is not divisible by 6. Sufficient.

(2) The greatest common factor of z and 15 is 3 --> if z=3 then the answer is NO but if z=6 then the answer is YES. Two different answers, not sufficient.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

I am not clear with this question.First statement tells us that GCF of z & 12 is 3,which indicates that z is atleast 3 or in other words is a multiple of 3. IF z is a multiple of 3,then it may or may not be divisible by 6.For e.g 3,9,27 is not divisible by 6 but 18 and 6 are divisible by 6.So how can we say that Statement 1 is sufficient???

I am not clear with this question.First statement tells us that GCF of z & 12 is 3,which indicates that z is atleast 3 or in other words is a multiple of 3. IF z is a multiple of 3,then it may or may not be divisible by 6.For e.g 3,9,27 is not divisible by 6 but 18 and 6 are divisible by 6.So how can we say that Statement 1 is sufficient???

This is explained in the post just above yours: if z were divisible by 6 (for example 6, 12, 18, ...) then the GCF of z and 12 (which is also divisible by 6) would have been more than 3 (6 or 12) and since the GCF is 3 then z is not divisible by 6. Sufficient.
_________________

Bunuel,I had read your post but I am not clear as to how we can rule out that Z can be a multiple of 3 that may or may not be divisible by 6.Does GCF define the divisibility of z when we dont know which factors z has other than 3

Bunuel,I had read your post but I am not clear as to how we can rule out that Z can be a multiple of 3 that may or may not be divisible by 6.Does GCF define the divisibility of z when we dont know which factors z has other than 3

Again, z cannot be divisible by 6, because if it were divisible by 6 then the greatest common factor of z and 12 would have been 6 (or 12) not 3. For example if z=6 then GCF of z=6 and 12 is 6 not 3.
_________________

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...