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

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Joined: 18 Apr 2015
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Kudos [?]: 972 [3] , given: 4479

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

10% (01:20) correct 90% (02:02) wrong based on 10 sessions

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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Director
Joined: 03 Sep 2017
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Kudos [?]: 327 [2] , given: 66

Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  16 Oct 2017, 00:54
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KUDOS
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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Joined: 24 Sep 2018
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Kudos [?]: 0 [0], given: 0

Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  26 Sep 2018, 11:30
i can't understand this one.. can someone else offer a different explanation?
Moderator
Joined: 18 Apr 2015
Posts: 4876
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Kudos [?]: 972 [0], given: 4479

Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  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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