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Math Expert V
Joined: 02 Sep 2009
Posts: 55193
b, c, and d are consecutive even integers such that 2 < b < c < d. Wha  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 35% (01:55) correct 65% (02:04) wrong based on 78 sessions

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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

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Posts: 1352
Concentration: Accounting, Finance
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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha  [#permalink]

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

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3
1
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
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Manager  S
Joined: 04 Jun 2018
Posts: 157
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]

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three consecutive integers are: 2k-2, 2k, 2k+2
if you take 2 out from each of the numbers:
8(k-1)(k)(k+1)

Also according to the question:
2k-2>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]

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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
Intern  B
Joined: 27 Jun 2015
Posts: 26
GRE 1: Q158 V143 Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha  [#permalink]

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Ans should be E as b ,c,d can take vakues 14,16,18 respectively and product of them is divisible by 96

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

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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
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Intern  B
Joined: 19 Feb 2019
Posts: 10
Concentration: Marketing, Statistics
Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha  [#permalink]

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1
1
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  B
Joined: 05 Mar 2018
Posts: 25
Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha  [#permalink]

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1
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  B
Joined: 19 Feb 2019
Posts: 10
Concentration: Marketing, Statistics
Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha  [#permalink]

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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  B
Joined: 26 Dec 2018
Posts: 3
Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha  [#permalink]

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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, 19:18
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# b, c, and d are consecutive even integers such that 2 < b < c < d. Wha

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