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a, b, c, and d are positive consecutive integers and a < b <
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17 Aug 2013, 09:34
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a, b, c, and d are positive consecutive integers and a < b < c < d. If the product of b, c, and d is twice that of a, b, and c, then bc = A 2 B 6 C 12 D 20 E 30
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Re: a, b, c, and d are positive consecutive integers and a < b <
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17 Aug 2013, 09:44
Stiv wrote: a, b, c, and d are positive consecutive integers and a < b < c < d. If the product of b, c, and d is twice that of a, b, and c, then bc = A 2 B 6 C 12 D 20 E 30 a, b, c, and d are positive consecutive integers since a b c d are consecutive integers
therefore \(d= a+3\) now given \(2abc = bcd\) \(bc(2ad) = 0\) since they are positive hence bc cant be equal to 0therefore \(2ad = 0\) putting \(d = a+3\) \(2aa3=0.\) \(a=3\) therefore \(b= 4 and c= 5\) hence bc = \(4*5 = 20\) HENCE D
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Re: a, b, c, and d are positive consecutive integers and a < b <
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17 Aug 2013, 09:51
Stiv wrote: a, b, c, and d are positive consecutive integers and a < b < c < d. If the product of b, c, and d is twice that of a, b, and c, then bc = A 2 B 6 C 12 D 20 E 30 Here, bcd = 2 abc , so, d = 2a and we know, a<b<c<d so the series is like, 3<4<5<6 Finally, bc = 4 × 5 = 20
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Re: a, b, c, and d are positive consecutive integers and a < b <
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10 Apr 2018, 10:44
Bunuel niks18 chetan2uQuote: a, b, c, and d are positive consecutive integers and a < b < c < d. If the product of b, c, and d is twice that of a, b, and c, then bc = A 2 B 6 C 12 D 20 E 30 My sol matched with blueseas let me know if there is any better approach?
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Re: a, b, c, and d are positive consecutive integers and a < b <
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10 Apr 2018, 13:40
Asifpirlo wrote: Stiv wrote: a, b, c, and d are positive consecutive integers and a < b < c < d. If the product of b, c, and d is twice that of a, b, and c, then bc = A 2 B 6 C 12 D 20 E 30 Here, bcd = 2 abc , so, d = 2a and we know, a<b<c<d so the series is like, 3<4<5<6 Finally, bc = 4 × 5 = 20 I did the problem this way and got the correct answer, but I think the primary reason why this approach worked is because we are told that the variables were positive integers. I don't think this approach would have worked if the sign wasn't mentioned.



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a, b, c, and d are positive consecutive integers and a < b <
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10 Apr 2018, 17:54
adkikani wrote: Bunuel niks18 chetan2uQuote: a, b, c, and d are positive consecutive integers and a < b < c < d. If the product of b, c, and d is twice that of a, b, and c, then bc = A 2 B 6 C 12 D 20 E 30 My sol matched with blueseas let me know if there is any better approach? Hi adkikani, I am not sure about "Better" but yes there are other approaches to solve the problem. Now it is also given that \(2abc=bcd =>d=2a\). You can work backwards through options  2=1*2, hence b=1,c=2,d=3 and a=0, which is not possible as a is positive. 6=2*3, hence b=2,c=3,d=4 and a=1, but d=2a, Hence ignore 12=3*4, hence b=3,c=4,d=5 and a=2, again not possible 20=4*5, hence b=4, c=5, d=6 and a=3. here d=2a. Hence our answer 30=5*6, hence b=5,c=6,d=7 and a=4, which is not possible



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Re: a, b, c, and d are positive consecutive integers and a < b <
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10 Apr 2018, 18:26
niks18Quote: By looking at the answer choices, you should realize that both\(b\) & \(d\) has to be Even. Can you verify highlighted text again? Fundamentally, EITHER of two numbers is suff for product to be EVEN. Is your rationale based on the fact that a,b,c,d are consecutive integers?
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a, b, c, and d are positive consecutive integers and a < b <
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10 Apr 2018, 18:50
adkikani wrote: niks18Quote: By looking at the answer choices, you should realize that both\(b\) & \(d\) has to be Even. Can you verify highlighted text again? Fundamentally, EITHER of two numbers is suff for product to be EVEN. Is your rationale based on the fact that a,b,c,d are consecutive integers? Hi adkikaniMy bad, I read the question as b*d instead of b*c. Yes you are correct, in your observation. I have edited my solution




a, b, c, and d are positive consecutive integers and a < b < &nbs
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