xy13-4-9

Rachmaninoff

xy13-4-9

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?

 A:1/4 B:3/8 C:1/2 D:5/8 E:3/4

 A:1/4 B:3/8 C:1/2 D:5/8 E:3/4

himba

这个题分两种情况, 如果看到三数连乘的题一般情况下也都是这样分类的

1. 第一个数是偶数,

那第二个数肯定是奇数咯, 第三个数是偶数, 所以三数中两数都是偶数, 而且至少可以被八整除, 因为第三个数至少可以被四整除.(这些规律做gmat题的时候是不能想的,脱口而出,因为时间太紧迫了),这种情况中n就是任意偶数，96/2=48，共48个偶数

2. 第一个数是奇数

那第二个数就是偶数了,第三个数就是奇数了,这三个数当中有一个数是偶数,要想被八整除前提是中间那个偶数能被八整除. 96/8=12， 共12个

那算概率吧，一共96个数，（48+12）/96

Rachmaninoff

还是不怎么理解呀！！

luolihua73

(48+12)/96=5/8

any even number will come out with at least 8 because the least even number will have 2*(2+1)*(2+2) which is divisible by 8. 
any odd number that's 1 less than multiple of 8 will count also.
no other odd number will be divisible by 8 because only even will be (n+1) and will not be multiple of 8.