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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<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.