Great explanation, krishnasty (but you're missing one key component that I'll explain in a second).
One point worth making - on these pattern-based problems, it can be really helpful to spend a little time thinking about why that pattern holds. If you look at the pattern you established, if n is even the product of n, n+1, and n+2 is divisible by 8, but if n is odd it's not. Why?
Well, every second number is even, so you're guaranteed to get Even, Odd, Even if n is even. And since every second even number is a multiple not just of 2 but also of 4 (2, 4, 6, 8, 10, 12, 14, 16...) then if you start with an even number you're bound to get a pairing of 2 * 4. So, indeed, every even n will give you a product divisible by 2, and you can prove that using properties of numbers.
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This problem is a little trickier than just that, too. Keep in mind that, above, we were manufacturing 8s out of a multiple of 2 and a multiple of 4. But there are some natural 8s in this set, too, since we're using all the numbers from 1 to 96. Sure, 2*3*4 works as does 6*7*8. But what about n = 15? 15*16*17 will also work since 16 is a natural multiple of 8 just on its own. And we need to account for those. Surrounding the number 16, it will be involved in:
n = 14 (14*15*16)
n = 15 (15*16*17)
n = 16 (16*17*18)
Well, we've already counted those for 14 and 16 since we're using all the even values of n already. But we haven't accounted for the possibility that n is odd and n+1 is a multiple of 8. That will happen at n = 7, 15, 23, 31... Or you can just look at the multiples of 8 in that set - 96/8 = 12, so there are 12 additional values of n that will still give us a product divisible by 8. So that gets us to the 48 even values of n + the 12 values of n+1 that give us an 8, so there are 60 out of the 96 possible values of n that will work, and we're up to 62.5% as the probability.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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