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

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b, c, and d are consecutive even integers such that 2 < b < [#permalink] New post 12 Aug 2017, 10:05
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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?

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[Reveal] Spoiler: OA
48

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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink] New post 16 Oct 2017, 00:54
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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.
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink] New post 26 Sep 2018, 11:30
i can't understand this one.. can someone else offer a different explanation?
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink] New post 27 Sep 2018, 09:28
Expert's post
Picking numbers

Consecutive even integers

2 < 4 < 6 < 8

4,6, and 8 actually are = \(2^6 * 3\)

The largest divisor of b*c*d is 48.

Hope this helps.
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Re: b, c, and d are consecutive even integers such that 2 < b <   [#permalink] 27 Sep 2018, 09:28
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b, c, and d are consecutive even integers such that 2 < b <

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