Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 61385

b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
28 Jan 2019, 00:28
Question Stats:
35% (01:57) correct 65% (02:03) wrong based on 85 sessions
HideShow timer Statistics
b, c, and d are consecutive even integers such that 2 < b < c < d. What is the largest positive integer that must be a divisor of bcd? A. 12 B. 16 C. 24 D. 48 E. 96
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5894
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
28 Jan 2019, 01:57
Bunuel wrote: b, c, and d are consecutive even integers such that 2 < b < c < d. What is the largest positive integer that must be a divisor of bcd?
A. 12 B. 16 C. 24 D. 48 E. 96 b, c and d are consecutive even integers such that 2<b<c<d. Let's assume b = 2n, then c = 2n+2 , d = 2n+4 and n>1 since 2<b<c<d. Then the expression bcd = 2n(2n+2)(2n+4)= 8n(n+1)(n+2). Now n(n+1)(n+2) is the product of 3 consecutive integers which is always divisible by 6. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48 IMO D




VP
Joined: 31 Oct 2013
Posts: 1491
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
28 Jan 2019, 01:54
Bunuel wrote: b, c, and d are consecutive even integers such that 2 < b < c < d. What is the largest positive integer that must be a divisor of bcd?
A. 12 B. 16 C. 24 D. 48 E. 96 Note: b, c, d are consecutive even integer**** let's put some values for b,c,d. b<c<d = 4<6<8. bcd = 72. 72 is not divisible by 96 , 48 and 16. we are left with 24 and 12. b=6 c=8 d=10 bcd = 6*8*10 = 480. divisible by both 12 and 24. So, 24 is our answer. Largest value that divide into any combination of bcd. C is the correct answer.



Current Student
Joined: 04 Jun 2018
Posts: 155
GMAT 1: 610 Q48 V25 GMAT 2: 690 Q50 V32 GMAT 3: 710 Q50 V36

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
28 Jan 2019, 20:57
three consecutive integers are: 2k2, 2k, 2k+2 if you take 2 out from each of the numbers: 8(k1)(k)(k+1)
Also according to the question: 2k2>2 2k>4 k>2
Now Lets take any values: 234 345 456 567(only divisible by 2*)
Every three consecutive numbers is divisible by at least 2* and every 3 consecutive integers is divisible by 3.
Hence the maximum divisor for every variable bcd: 2*3*8=48
Hence Ans D



Intern
Joined: 02 Jan 2019
Posts: 2

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
08 Apr 2019, 03:18
Note that either b, c or d will always be a multiple of 3, so answer must be a multiple of 3.
Smallest possible values for b, c and d are 3,4 and 5, so min. possible bcd is 60. Therefore, answer must be the the largest multiple of 3 below 60.
So Answer D: 48



Manager
Joined: 27 Jun 2015
Posts: 68

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
08 Apr 2019, 03:51
Ans should be E as b ,c,d can take vakues 14,16,18 respectively and product of them is divisible by 96
Posted from my mobile device



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9432
Location: United States (CA)

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
10 Apr 2019, 16:52
Bunuel wrote: b, c, and d are consecutive even integers such that 2 < b < c < d. What is the largest positive integer that must be a divisor of bcd?
A. 12 B. 16 C. 24 D. 48 E. 96 We can let b = 2k for some integer k ≥ 2. So c = 2k + 2 and d = 2k + 4. Therefore, we have bcd = 2k(2k + 2)(2k + 4) = 2k(2(k + 1))(2(k + 2)) = 8k(k + 1)(k + 2) Notice that k(k + 1)(k + 2) is a product of 3 consecutive integers and thus it’s divisible by 3! = 6. Thus. 8k(k + 1)(k + 2) must be divisible by 8(6) = 48. Answer: D
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Manager
Joined: 19 Feb 2019
Posts: 114
Location: India
Concentration: Marketing, Statistics
GPA: 3

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
25 Apr 2019, 16:16
How can it be 48? they have asked for 3 consecutive even integers greater than 2. Won't the largest positive integer which is a divisor of these three even integers be 96? Taking 4,6 and 8 as an e: the product is 192 so the largest integer which is the divisor should be 96 right?
Am I missing something here? Pls help



Intern
Joined: 05 Mar 2018
Posts: 28

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
01 May 2019, 01:19
devavrat wrote: How can it be 48? they have asked for 3 consecutive even integers greater than 2. Won't the largest positive integer which is a divisor of these three even integers be 96? Taking 4,6 and 8 as an e: the product is 192 so the largest integer which is the divisor should be 96 right?
Am I missing something here? Pls help if you take 4, 6, 8 ==> 96 ( but also 48) try the next set ==> 6 * 8 * 10 = 480 .. which is not divisible by 96. next set ==> 8 * 10 * 12 = 960 .. divisible by 96 ( but also 48) and the pattern continues. So, any set of bcd is definitely divisible by 48.



Manager
Joined: 19 Feb 2019
Posts: 114
Location: India
Concentration: Marketing, Statistics
GPA: 3

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
01 May 2019, 02:14
480 is divisible by 96. 96 is not divisible by 10X12X14 Then its is not divisible by 18X20X22 Whereas as all the numbers are divisible by 48 MeBossBaby wrote: devavrat wrote: How can it be 48? they have asked for 3 consecutive even integers greater than 2. Won't the largest positive integer which is a divisor of these three even integers be 96? Taking 4,6 and 8 as an e: the product is 192 so the largest integer which is the divisor should be 96 right?
Am I missing something here? Pls help if you take 4, 6, 8 ==> 96 ( but also 48) try the next set ==> 6 * 8 * 10 = 480 .. which is not divisible by 96. next set ==> 8 * 10 * 12 = 960 .. divisible by 96 ( but also 48) and the pattern continues. So, any set of bcd is definitely divisible by 48.



Intern
Joined: 26 Dec 2018
Posts: 17

Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
Show Tags
01 May 2019, 18:18
MeBossBaby wrote: devavrat wrote: How can it be 48? they have asked for 3 consecutive even integers greater than 2. Won't the largest positive integer which is a divisor of these three even integers be 96? Taking 4,6 and 8 as an e: the product is 192 so the largest integer which is the divisor should be 96 right?
Am I missing something here? Pls help if you take 4, 6, 8 ==> 96 ( but also 48) try the next set ==> 6 * 8 * 10 = 480 .. which is not divisible by 96. next set ==> 8 * 10 * 12 = 960 .. divisible by 96 ( but also 48) and the pattern continues. So, any set of bcd is definitely divisible by 48. Dude how is 480 not divisible by 96 ? 96×5 = 480 Posted from my mobile device




Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
[#permalink]
01 May 2019, 18:18






