Author 
Message 
TAGS:

Hide Tags

Director
Joined: 09 Aug 2006
Posts: 517

If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
Updated on: 16 Feb 2015, 03:19
1
This post received KUDOS
2
This post was BOOKMARKED
Question Stats:
86% (00:40) correct 14% (00:46) wrong based on 307 sessions
HideShow timer Statistics
If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following? (A) 10 (B) 12 (C) 14 (D) 16 (E) 18
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by Amit05 on 04 Jul 2007, 20:50.
Last edited by Bunuel on 16 Feb 2015, 03:19, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.



Director
Joined: 09 Aug 2006
Posts: 746

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
05 Jul 2007, 00:12
Amit05 wrote: If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following? (A) 10 (B) 12 (C) 14 (D) 16 (E) 18
Though I got the correct ans by brute force. Just wondering which math rule could be applied here.
The way I figure it is taking the product of the prime factors which are not common among the two.
6= 3*2
8= 2*2*2
prod. of prime factors not common = 2*2*3=12
Answer B



Director
Joined: 15 Jun 2007
Posts: 503

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
05 Jul 2007, 00:13
1
This post received KUDOS
2
This post was BOOKMARKED
Find the prime factors of both, then multiply.
Ex)
6=(2)(3)
8=(2)(2)(2)
LCM = 24
24 is only divisible by 12. Answer is B.



CIO
Joined: 09 Mar 2003
Posts: 460

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
Updated on: 05 Jul 2007, 08:12
GK_Gmat wrote: Amit05 wrote: If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following? (A) 10 (B) 12 (C) 14 (D) 16 (E) 18
Though I got the correct ans by brute force. Just wondering which math rule could be applied here. The way I figure it is taking the product of the prime factors which are not common among the two. 6= 3*2 8= 2*2*2 prod. of prime factors not common = 2*2*3=12 Answer B
I like this explanation. If it's divisible by 6, it must have every prime 6 has, and the same thing is true for 8. So there must be 2,2,3. Even if you don't find the lowest common multiple, as in above, you should still just look for the answers with just some 2's and one 3. 12 is the only one.
Note that 18 doesn't work because there's an extra 3 in it.
Originally posted by ian7777 on 05 Jul 2007, 06:07.
Last edited by ian7777 on 05 Jul 2007, 08:12, edited 1 time in total.



Director
Joined: 16 May 2007
Posts: 539

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
05 Jul 2007, 07:17
I would go with the LCM approach since i believe it is more feasible when dealing with unusual numbers.



Senior Manager
Joined: 04 Jun 2007
Posts: 359

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
05 Jul 2007, 08:11
1
This post received KUDOS
METHOD 1
6
0, 6, 12, 18, 24
8
0, 8, 16, 24
LCM = 24 which is divisible by 12.
hence, B.
METHOD 2
I also like the factoring approach for smaller numbers.
6= 2.3
8= 2.2.2
therefore, 2.2.3= 12, hence, B.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11663
Location: United States (CA)
GRE 1: 340 Q170 V170

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
14 Feb 2015, 21:00
Hi All, In these sorts of situations, primefactorization is a great technical approach to get to the correct answer. There is another method that is actually pretty easy though  we can TEST VALUES. Since the prompt asks for what N MUST be divisible by, we just need to start with the SMALLEST positive integer for N that is DIVISIBLE by BOTH 6 and 8. Many Test Takers would say that 48 is the smallest integer, but it's NOT. The smallest integer is actually 24. Here's proof that's fairly easy to put together.... Multiples of 6: 6, 12, 18, 24 Multiples of 8: 8, 16, 24 Now, which of the answer choices divides into 24? Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************



Senior Manager
Joined: 20 Aug 2015
Posts: 392
Location: India

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
25 Feb 2016, 00:51
Amit05 wrote: If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following?
(A) 10 (B) 12 (C) 14 (D) 16 (E) 18 This can be easily solved by testing the values. Assume the number n = 24 (divisible by both 6 and 8) Of the options, Only Option B satisfies. Correct Option: B



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2602
GRE 1: 323 Q169 V154

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
14 Mar 2016, 00:49



Intern
Status: It's time...
Joined: 29 Jun 2015
Posts: 7
Location: India
Concentration: Operations, General Management
GPA: 3.59

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
27 May 2016, 05:45
1
This post received KUDOS
Amit05 wrote: If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following?
(A) 10 (B) 12 (C) 14 (D) 16 (E) 18 A very straightforward method involving the LCM exists to solve the problem. First of all, "n" is divisible by both 6 and 8; This implies "n" is a Multiple of both 6 and 8. Finding the LCM helps as all other Multiples of 6 and 8 may have extra prime factors which may lead to "n" be divisible by other numbers. e.g. the LCM of 6 and 8 is 24, but if we find any multiple of 6 and 8 such as 48 or 72 then apart from 12, the number 48 is also divisible by 16 and 72 by 18. Finally, check the divisiblity of the LCM by the options provided. Only one value should satisfy the divisibility condition. Cheers!!!... Press "Kudos" if you liked the explanation.
_________________
Perseverance, the answer to all our woes. Amen.



Intern
Joined: 04 Jun 2017
Posts: 8

If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
02 Sep 2017, 08:57
n is divisible by both 6 and 8 means .... n = 6*8
n = 6*8 = (2^4) * 3 implies that the answer must meet two conditions: (1) be divisible by the product of its prime which is 6=2*3. This by itself narrows down the answer to B=12 or E=18. (2) contains at most one 3 and at most four 2's. E is not correct because it factors into two 3's which is one too many 3's.
The answer is B.



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2623
Location: United States (CA)

Re: If an integer n is divisible by both 6 and 8, then it must [#permalink]
Show Tags
07 Sep 2017, 07:10
Amit05 wrote: If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following?
(A) 10 (B) 12 (C) 14 (D) 16 (E) 18 Let’s first find the LCM of 6 and 8. The prime factorization of 6 is 3 x 2 and the prime factorization of 8 is 2 x 2 x 2. Thus, the LCM of 6 and 8 is 3 x 2 x 2 x 2 = 24. Of all the answer choices, only 12 is a factor of 24, so n is divisible by 12. Answer: B
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: If an integer n is divisible by both 6 and 8, then it must
[#permalink]
07 Sep 2017, 07:10






