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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]  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?

[Reveal] Spoiler: OA
48

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Kudos [?]: 217 [0], given: 66

Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  16 Oct 2017, 00:54
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.
Re: b, c, and d are consecutive even integers such that 2 < b <   [#permalink] 16 Oct 2017, 00:54
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# b, c, and d are consecutive even integers such that 2 < b <

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