Is the integer z divisible by 6?

My sincere apologies. I got it now. GCD will be the multiplication of common factors. So it will be 6 for 12 & 12.

Thanks for all your help.

For 12 & 12; GCD is 12.

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.

its NOT UNIQUE common factors but ALL COMMON FACTORS. So, in above common factos are 2, 2 and 3, hence 12.

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.

This question is from MGMAT Number Properties guide buddy and it says

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.

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

GCF(630,240)=3^1*2^1*5^1=30
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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
***********************************************

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
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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.

Answer is A.

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

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.

z must be divisible by 6
z must be divisible by 2,3


St1: Hcf(z,12)=3
\(12=2^2*3\)
\(z=3*x\)

HCF= 3 in z implies x can have no even number no two's Hence Z is not div by 6
Suff

St2: Hcf(z,15)=3
\(15=3*5\\
z=3*x\)

HCF=3 in z implies x cannot be 5 but anything, so x may be odd or even
Not Suff

Option A

Can you give more sum examples of this type please?

gcf(z,12)= 3 that means z is not divisible by 6=> sufficient
gcf(z,15)=3 does not provide info of whether it is divisible by 3x2=>insufficient

option A