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标题: help for GWD4_9 [打印本页]

作者: kenlily    时间: 2006-1-21 13:38     标题: help for GWD4_9

Although I gussed the right answer(D),I don't konw how to get it!Pls. help for it.Thanks a lot!

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. ATH connecttype="rect" gradientshapeok="t" extrusionok="f">ATH>1/4

B. 3/8

C. 1/2

D. 5/8

E. 3/4


作者: mjiirggh    时间: 2006-1-22 13:02

当n是偶数的时候:

  设n=2k,n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(k+1)(2k+1)

  注意k和k+1两者必有一个是偶数,所以4k(k+1)(2k+1)必定能被8整除。即满足条件的n有96/2=48个。

当n是奇数的时候:

  注意到n(n+1)(n+2)中只有n+1是偶数,故只有当n+1能被8整除才能满足题目条件。n的取值可以是7,15,23……共  12个。

综上所述,n的取值一共有48+12=60个,概率是60/96=5/8。


作者: kenlily    时间: 2006-1-22 19:45

many   thanks!!!!!!!!!!!




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