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Is the integer Z divisible by 6? 1. The GCF of z and 12 is [#permalink]
25 Jun 2011, 11:22

00:00

A

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Difficulty:

55% (hard)

Question Stats:

33% (01:33) correct
67% (01:16) wrong based on 12 sessions

Is the integer Z divisible by 6?

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

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?

Re: Number properties - DS [#permalink]
25 Jun 2011, 11:28

Is the integer Z divisible by 6?

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.

Re: Number properties - DS [#permalink]
25 Jun 2011, 11:42

Thanks. But GCD is greatest common divisior or factor (GCF). So if z is 12 then also the GCF will be 3. Or I am not getting it correctly? _________________

Re: Number properties - DS [#permalink]
25 Jun 2011, 12:10

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? _________________

Re: Number properties - DS [#permalink]
25 Jun 2011, 12:22

enigma123 wrote:

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. _________________

Re: Number properties - DS [#permalink]
25 Jun 2011, 12:37

enigma123 wrote:

Ok - for GCD of 12 & 12

12 - 2,2,3 12- 2,2,3

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 **************************************** _________________

Re: Number properties - DS [#permalink]
25 Jun 2011, 17:57

1

This post received KUDOS

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.