If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being
divisible by 8?
A 25%
B 50%
C 62.5%
D 72.5%
E 75%
divisible by 8?
A 25%
B 50%
C 62.5%
D 72.5%
E 75%
First the solution... then the explanation. Without the explanation, the solution has only limited value to your preparation.
For any number to be divisible by 8, it must be divisible by 2 - three times. (2*2*2=8)
n, n+1, and n+2 are consecutive integers, which means that we have either two even numbers and an odd one (when n is even) or two odds and an even (when n is odd).
We know that no odd number is divisible by 2, meaning out of all the n's that are odd, only those whose n+1 is divisibly by 8 qualify. For example, if n=7, then we have 7*8*9, which is divisible by 8. Since 96/8 = 12, there are twelve n's that work in this case (7, 15, 23, ..., 95)
For the even n's, we still need the three 2's, but they can be distributed among both even numbers. Fortunately, with two consecutive even numbers, one of the even's is always divisible by 4 (two 2's) and the other by a single 2. This means ALL even n's qualify. So 96/2 = 48.
Total, we have 48+12 = 60 qualifying n's out of 96 total, which represents 60/96 = 10/16 = 5/8 = 62.5%
Our answer is C.
Now, the explanation...
What are the key insights here?
1) Understand that the number 8 is misleading, we should be looking at primes.
2) Understand that the complex algebra (how many of you multiplied it out??) distracts you from the idea of consecutive integers.
The solution path consists of prompting yourself to see the problem differently, on your own terms. This means seeking out examples, plugging in numbers, and understanding the nature of the problem. This will lead you to the most efficient solution path.
If you're not cultivating your process while you prep and focusing on the math/language rules/answers, you're wasting your time.
Ben
