Algebra/Probability

[light_speed]

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

I have calculated that there are 57 numbers that are divisible by 8 from the above forumula, but dividing 57/96 to give me an answer is not working. Where am i going wrong?

Thanks!

[simplyjat]

The answer is definitely not 57. You have to approach these kind of questions differently. You have to identify the pattern. The pattern here is for a multiple of 8 the range 1-8 is similar to 1-96. So you can just concentrate on x between 1-8 and get the desired results.

List down n(n + 1)(n + 2) for all number from 1 to 8
1> 1.2.3 - not divisible by 8
2> 2.3.4 - divisible by 8
3> 3.4.5 - not divisible by 8
4> 4.5.6 - divisible by 8
5> 5.6.7 - not divisible by 8
6> 6.7.8 - divisible by 8
7> 7.8.9 - divisible by 8
8> 8.9.10 - divisible by 8

so we see when x = 2, 4, 6, 7, 8 the sum is divisible is 8. so the answer is 5/8....

Why the answer is same as that of 1-96 is left as an exercise

[tomato1]

There is a rule that product of two consecutive even no's is multiple of 8.

In this qus n and (n+2) will be consecutive even integers for even values of n. Since there can be 48 even values of n. so n(n+2) will be a multiple
of 8 for 48 values of n.

the product n(n+1)(n+2) can be divisible by 8 in two ways....
1. when product of n(n+2) is multiple of 8. As said above for 48 values of n.


2. when (n+1) is multiple of 8 which is possible when n =7, 15, .........95. that is for 12 values of n.


so n(n+1)(n+2) can be divisible by 8 for 48+12 = 60 values.


thus prob=60/96=5/8

[humeixia]