If a is an integer from 1 to 96, what is the probability for a*(a+1)*(a+2) being divisible by 8?	Choose one:
A) a. 25%
B) b. 50%
C) c. 62.5%
D) d. 72.5%
E) e. 75%
There is another way I approach such questions if I am short of time (otherwise I prefer the logical approach given above) - Brute Force/ Pattern Recognition/Intuition - whatever you may want to call it.   We need to find the numbers in which the product is a multiple of 8. I know I get a multiple of 8 after every 8 numbers. I will also get an 8 when I multiply 4 by an even number. In first 8 numbers, I have exactly two multiples of 4.   Basically, I figure that I should look at the first 8 cases. In all other cases, the pattern will be repeated. It helps that n can be from 1 to 96 i.e. a multiple of 8:   1*2*3 N   2*3*4 Y   3*4*5 N   4*5*6 Y   5*6*7 N   6*7*8 Y   7*8*9 Y   8*9*10 Y   5 of the first 8 products are divisible by 8 so my answer would be 5/8 = 62.5% c. 62.5%
<result>C</result>
If a is an integer from 1 to 96, what is the probability for a*(a+1)*(a+2) being divisible by 8?