标题: gwd4-9 [打印本页]

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作者: qwrersaa    时间: 2004-9-7 12:27     标题: gwd4-9

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?

key: 5/8

how to calculate?

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作者: ppp1    时间: 2004-9-7 12:34

1,当n为偶数时,n(n+1)(n+2)显然可以被8整除,因为n是2的倍数时,n+2是4的倍数, 这种情况一共有96/2=48种;

2, 还有另一种情况是n为奇数,但是n+1是8的倍数, 这种情况一共有96/8=12种;

一共可能的组合有96种, 因此,答案为 (48+12)/96=5/8

不知道说清楚了没有..........

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