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# If an integer n is to be chosen at random from the integers

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Joined: 07 Nov 2004
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Kudos [?]: 20 [0], given: 0

If an integer n is to be chosen at random from the integers [#permalink]  02 Feb 2005, 09:55
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If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
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Joined: 03 Jan 2005
Posts: 2250
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Kudos [?]: 216 [0], given: 0

n=4k+1, n(n+1)(n+2)=(4k+1)(4k+2)(4k+3)=2(4k+1)(2k+1)(4k+3) can bedivided by 2, but not 8
n=4k+2, n(n+1)(n+2)=(4k+2)(4k+3)(4k+4)=8(2k+1)(4k+3)(k+1) divisable by 8
n=4k+3, n(n+1)(n+2)=(4k+3)(4k+4)(4k+5)=4(4k+3)(k+1)(4k+5)
If k+1 is even then divisible by 8, otherwise, divisible by 4, not 8
n=4k+4, n(n+1)(n+2)=(4k+4)(4k+5)(4k+6)=8(k+1)(4k+5)(2k+3), divisible by 8

Therefore 1/4+1/4+1/8=5/8
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