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b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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28 Jan 2019, 01:28
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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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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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28 Jan 2019, 02:57
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




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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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25 Apr 2019, 17:16
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



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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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01 May 2019, 02:19
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.



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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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28 Jan 2019, 02:54
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.



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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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28 Jan 2019, 21:57
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



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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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08 Apr 2019, 04:18
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



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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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08 Apr 2019, 04:51
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
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10 Apr 2019, 17:52
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
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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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01 May 2019, 03:14
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.



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Re: b, c, and d are consecutive even integers such that 2 < b < c < d. Wha
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01 May 2019, 19:18
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
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