) Find number that are divisible by 8 btw 1 and 96 inclusive: 8 x 12 =96. So 12 numbers.duongthang wrote:if an integer n is to be chosen at random from the intergers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8
1/4
3/8
1/2
5/8
3/4
pls, help with this
Examine n(n+1)(n+2)
Notice if n is divisible by 8, then n+1 and n+2 will not be and when n+1 is divisible by 8, n+2 and n will not be and vice versa. So you have a case of mutually exclusive possibilities.
By the Fundamental Counting Principle in such cases we add the number of outcomes
For N we have 12 from above
For N+1: (N+1)/8=q if N = 8q-1. There are q =12 numbers that satisfy this.
For N+2: (N+2)/8=p if N = 8p-2 There are p=12 numbers that satisfy this also.
Since these are the only possibilities ( see that 8 x8 x 8 not possible) we add:
12 +12+12=36 numbers that are divisible by 8 between 1 and 96 inclusive
Prob 36/96=3/8
Choose B
Tools Needed for quick solution
1) An understanding that 8 x 12 = 96 means there are 12 numbers divisible by 8 between 1 and 96 inclusive
2)An understanding of Fundamental Counting Principle when Mutually Exclusive Outcomes are involved
3) Transforming the concept of divisibility into math to easily get how many of n+1 or n+2 are divisble by 8 within a specified range: (N+1)/8=q if N = 8q-1. This is not too diffrent from 8 x 12, except that we can just easily substitute 1, 2, 3....12 to try to get to 96.
4) And of course plain old probability which is really trivial here.
If these can occur easily to you without putting pen on paper, more power to YOu!!! Mind you these beasts would still appear in another formidable guise.
