b, c, and d are consecutive even integers such that 2

Question

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

Archit3110 · Accepted Answer

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

KSBGC · Answer

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.

nitesh50 · Answer

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

Guillemck · Answer

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

vishumangal · Answer

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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ScottTargetTestPrep · Answer

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

devavrat · Answer

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

MeBossBaby · Answer

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.

devavrat · Answer

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

ab965 · Answer

Dude how is 480 not divisible by 96 ? 96×5 = 480

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Alvarito · Answer

i think there is something missing out.

at first, I thought it was 2<b(4)<c(6)<d(8), so the product of BCD would be 192 which is divisible by 96, but then, the right answer is 48. I was reading other answers but still don't get it, the problem never says the numbers after 2 are 4,6 and 8, they could be anyone, any number,

the problem is, I can test different numbers, but sometimes 96 is the right answer, sometimes no.
am I doing something wrong? can someone please help out?

GAURI2828 · Answer

how is product of 4 6 8 = 72???? its 192

BrushMyQuant · Answer

Given that b, c, and d are consecutive even integers such that 2 < b < c < d and we need to find What is the largest positive integer that must be a divisor of bcd

As b, c and d are consecutive even numbers so one of them will be a multiple of 2, one will be a multiple of 4 and one will be a multiple of 6

Ex: 4, 6, 8. 4 -> multiple of 2 and 4, 6 -> multiple of 6, 8 -> multiple of 2 and 4

=> b*c*d will be a multiple of 2*4*6 = 48

So, Answer will be D
Hope it helps!

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b, c, and d are consecutive even integers such that 2 < b < c < d. Wha

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