GWD管卫东 GMAT 第8套 求讲解 先谢过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 A.1/4 B.3/8C.1/2D.5/8 E.3/4

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GWD管卫东 GMAT 第8套 求讲解 先谢过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 A.1/4 B.3/8C.1/2D.5/8 E.3/4
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GWD管卫东 GMAT 第8套 求讲解 先谢过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 A.1/4 B.3/8C.1/2D.5/8 E.3/4
GWD管卫东 GMAT 第8套 求讲解 先谢过
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
A.1/4
B.3/8
C.1/2
D.5/8
E.3/4

GWD管卫东 GMAT 第8套 求讲解 先谢过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 A.1/4 B.3/8C.1/2D.5/8 E.3/4
题目：整数n是1-96之间随机选出的数（包括1和96）,n（n+1）（n+2）除以8余数为0的概率是多少?
太简单了.只有8的倍数才能被8除尽.因此n（n+1）（n+2）必然是8的倍数.
基本思路有了,最简单的方法你可以穷举.
n可以是6,7,8,14,15,16,22,23,24,30,31,32,38,39,40,46,47,48,54,55,56,62,63,64,70,71,72,78,79,80,86,87,88,94,95,96
一共36个.而n可取范围为1-96,一共96个.因此概率为36/96=3/8,选B.
有技巧性的方法如下,n,（n+1）,（n+2）任何一个只要是8的倍数即可.因此n有96/8=12种,同理,n+1和n+2分别有12种.所以总共为12*3=36种.
同上,36/96=3/8,选B.
这是GMAT数学题里面相对来说简单的了,如果你的数学基础较差,建议花一个星期的时间去把这个月机经先做了；另外GWD数学题没有prep的好,可以直接入手prep数学.

依照楼上解释 那当n=10的时候，10*11*12是可以被8整除的，答案应该是D。

这道题的正确答案是D 不知道答案你们还瞎教人- -

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