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?
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Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610

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?
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610

For 12 & 12; GCD is 12.
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~fluke

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

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

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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
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~fluke

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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
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~fluke

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Thanks Spidy 001. You guys are too good. Really appreciate your help and everyone who helped me.
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610