Problem soving: Integer 1 to 96

[magauovazamat]

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

Answer is D.

I was confused and went in the wrong direction.

[RonPurewal]

divisibility by 8 is something that repeats every eight integers -- i.e., the pattern of divisibility is exactly the same in each cycle of 8 integers -- so you don't have to consider all the integers from 1 to 96.
1 to 96 is exactly twelve complete cycles of eight integers in a row. therefore, you can just look at 1 through 8 (one cycle). the probability will be the same, since everything will just happen the same way twelve times.

so ...
n = 1 --> 1 x 2 x 3, not divisible by 8
n = 2 --> 2 x 3 x 4, divisible by 8
n = 3 --> 3 x 4 x 5, not divisible by 8
n = 4 --> 4 x 5 x 6, divisible by 8
n = 5 --> 5 x 6 x 7, not divisible by 8
n = 6 --> 6 x 7 x 8, divisible by 8
n = 7 --> 7 x 8 x 9, divisible by 8
n = 8 --> 8 x 9 x 10, divisible by 8

that's 5 out of 8. this exact cycle will repeat itself every 8 integers, so the overall probability will also be 5 out of 8.

[magauovazamat]

Thank you!!! Get it!!!

[RonPurewal]

Thank you!!! Get it!!!

you're welcome.