[求助]GWD-08-10

bastencools

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                                                   答案是Ｄ，为什么呢？

SunnyApples

同意ls的 

bioliwu

1-96 INCLUSIVE

1) 连续数性质, 三个连续数相乘,中间为奇数,能被8 除.

这就有48个.从2*3*4---96*97*98

2)连续数中间为偶数8,或8的倍数.

7*8*9, 中间还有16,24,32,40,48,56，8K, 到,95*96*97,

共 12个

1)+2)为 60个.

一共有96个.

所以为 60/96=5/8

strawdog

when n is an even integer, n(n+1)(n+2) will definitely be divisible by 8, there are 48 even numbers in total.

when n is an odd integer, n(n+1)(n+2) will only be divisible by 8 when (n+1) is divisible by 8, there are 12 numbers in total.

(48+12)/96=5/8