December 10, 2018 December 10, 2018 10:00 PM PST 11:00 PM PST Practice the one most important Quant section  Integer properties, and rapidly improve your skills. December 11, 2018 December 11, 2018 09:00 PM EST 10:00 PM EST Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 475
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 11:22
Question Stats:
61% (01:10) correct 39% (01:17) wrong based on 360 sessions
HideShow timer Statistics
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?
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Senior Manager
Joined: 24 Mar 2011
Posts: 364
Location: Texas

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
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.
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



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 475
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 11:48
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.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Retired Moderator
Joined: 20 Dec 2010
Posts: 1820

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 11:49
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.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 475
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 11:53
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.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Senior Manager
Joined: 24 Mar 2011
Posts: 364
Location: Texas

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 12:04
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.



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 475
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
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?
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Retired Moderator
Joined: 20 Dec 2010
Posts: 1820

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
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.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 475
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 12:29
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.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Retired Moderator
Joined: 20 Dec 2010
Posts: 1820

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
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 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 ***********************************************
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Retired Moderator
Joined: 20 Dec 2010
Posts: 1820

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 12:46
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 ****************************************
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 01 Feb 2011
Posts: 660

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
25 Jun 2011, 17:57
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.



Math Expert
Joined: 02 Sep 2009
Posts: 51072

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
23 Feb 2012, 14:25



Intern
Joined: 16 Jul 2015
Posts: 40
GMAT 1: 580 Q37 V33 GMAT 2: 580 Q39 V31 GMAT 3: 560 Q40 V28 GMAT 4: 580 Q37 V32 GMAT 5: 680 Q45 V37 GMAT 6: 690 Q47 V37

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
17 Jun 2016, 08:54
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???



Math Expert
Joined: 02 Sep 2009
Posts: 51072

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
17 Jun 2016, 09:18
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 16 Jul 2015
Posts: 40
GMAT 1: 580 Q37 V33 GMAT 2: 580 Q39 V31 GMAT 3: 560 Q40 V28 GMAT 4: 580 Q37 V32 GMAT 5: 680 Q45 V37 GMAT 6: 690 Q47 V37

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
17 Jun 2016, 10:11
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



Math Expert
Joined: 02 Sep 2009
Posts: 51072

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
17 Jun 2016, 10:28



Manager
Joined: 17 Feb 2016
Posts: 91
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.12
WE: Education (Internet and New Media)

Is the integer z divisible by 6?
[#permalink]
Show Tags
23 Jun 2017, 00:15
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
_________________
Never stop fighting until you arrive at your destined place  that is, the unique you. Have an aim in life, continuously acquire knowledge, work hard, and have the perseverance to realise the great life. A. P. J. Abdul Kalam



NonHuman User
Joined: 09 Sep 2013
Posts: 9099

Re: Is the integer z divisible by 6?
[#permalink]
Show Tags
11 Aug 2018, 15:40
Hello from the GMAT Club BumpBot! 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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Is the integer z divisible by 6? &nbs
[#permalink]
11 Aug 2018, 15:40






