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标题: gwd2-9 [打印本页]

作者: Economics 时间: 2007-1-22 06:59 标题: gwd2-9

Q9:

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? ( KE: 5/8)

可以提供思路吗？谢谢~~

作者: goodfishs 时间: 2007-1-22 20:45

偶数全可以共48个, 奇数从7, 15…,95 共12个, (48+12)/96=5/8

作者: albertes 时间: 2007-1-25 06:34

1-96中有１２个８的倍数；１－９６中有４８个奇数（n+1）那么有４８个n使n(n+1)(n+2)可以被８整除，所以概率是（４８＋１２）／９６＝５／８

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