* Probability consect. integer divisible by 8

[WAN]

Dear folks, please review where is the mistake with the following probability problem:

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?

answ. said 5/8 but i got 1/8. where is the mistake?

n(n+1)(n+2) is consecutive integer and divisible by 8 is 8,16,24,32,40,48........96 (multiple number of 8)

thus, from 1 to 96 there are 12 number of 96 (12/96)--->1/8

thx

[pt]

You have to choose n from 1-96 and not the product value as I think you are trying to do.

From n=1-96, all even numbers will satisfy the condition that n(n+1)(n+2) be divisible by 8 (try it out). There are 48 such numbers (96/2). In addition, odd numbers n such that n=8x-1 (i.e. 1 less than a multiple of 8) will also satify the condition. There are 12 such numbers (96/8). Total = 60 values of n that satisfy the condition.

Answer is 60/96 = 5/8.

[WAN]

pt,

thx for your explanation...
i've got it.

[StaceyKoprince]

Please cite the source. If you don't, we will have to delete the entire problem. (Please don't forget about the rules, guys - the more time we have to spend making posts like this one, the less time we can spend answering test questions! If you haven't read the stickies or if it's been a while since you did... go read them again right now.)

[mann.pal7789]

source is paper test stacy..
stacy ur posts are just awesome.really easy to understand..thanks..i will definately meet when i come to us..

[tim]

What makes you think Stacey is in the US? :)