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Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 12:22
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Is the integer z divisible by 6? (1) The greatest common factor of z and 12 is 3. (2) The greatest common factor of z and 15 is 3. 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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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 12:28
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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.
st1 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.
so st1 is sufficient. A



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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 12:48
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.
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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 12:49
enigma123 wrote: 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.
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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 12:53
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.
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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 13:04
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. its NOT UNIQUE common factors but ALL COMMON FACTORS. So, in above common factos are 2, 2 and 3, hence 12.



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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 13:10
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?
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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 13:22
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.
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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 13:29
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.
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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 13:37
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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 GCF(630,240)=3^1*2^1*5^1=30 ************************************ 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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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 13:46
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enigma123 wrote: 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. 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 ****************************************
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Re: Is the integer z divisible by 6? [#permalink]
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25 Jun 2011, 18:57
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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.



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Re: Is the integer z divisible by 6? [#permalink]
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Re: Is the integer z divisible by 6? [#permalink]
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Re: Is the integer z divisible by 6? [#permalink]
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17 Jun 2016, 09:54
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???



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Re: Is the integer z divisible by 6? [#permalink]
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17 Jun 2016, 10:18
bhamini1 wrote: 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.
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Re: Is the integer z divisible by 6? [#permalink]
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17 Jun 2016, 11:11
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



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Re: Is the integer z divisible by 6? [#permalink]
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Is the integer z divisible by 6? [#permalink]
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23 Jun 2017, 01:15
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 SuffSt2: 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 SuffOption A
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Is the integer z divisible by 6?
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