IMO [spoiler] 50% [/spoiler]
i just plugged in few numbers for the value of n
n = 1 ....1*2*3 = 6 (not divisible)
n = 2 ....2*3*4 = 24 (divisible)
n = 3 ....3*4*5 = 60 (not divisible)
n = 4.....4*5*6 = 120 (divisible)
....
if you observe the pattern, every alternate solution is divisible by 8.
hence, my ans.. [spoiler] 50% [/spoiler]
Please help
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krishnasty
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Brian@VeritasPrep
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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.
______________________________________________________
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.
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.
______________________________________________________
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
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Thanks for your explanation Brian. I have a question though; when you consider the triplets( Set of three consecutive numbers) which have n+1 as a multiple of 8, how can we consider n+1=96 as the set is 1 through 96(inclusive). So the last triplet that has the middle term as a multiple of 8 would 87,88,89... Please explain.
Thanks!
Thanks!
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The only requirement here is that n ITSELF be between 1 and 96, inclusive.roshin wrote:Thanks for your explanation Brian. I have a question though; when you consider the triplets( Set of three consecutive numbers) which have n+1 as a multiple of 8, how can we consider n+1=96 as the set is 1 through 96(inclusive). So the last triplet that has the middle term as a multiple of 8 would 87,88,89... Please explain.
Thanks!
It is perfectly fine if n+1 and n+2 are greater than 96.
If n=95, then n(n+1)(n+2) = 95*96*97, in which case the middle factor is a multiple of 8.
Here's my solution:
Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8?
25%
50%
62.5%
72.5%
75%
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
Total number of even integers = 96/2 = 48.
Case 2: n(n+1)(n+2) = (odd)(multiple of 8)(odd):
The product will be a multiple of 8 if n+1 is a multiple of 8.
Between 1 and 96, the total number of multiples of 8 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying that there are 12 more favorable choices for n.
Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8 = 62.5%.
The correct answer is C.
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As a tutor, I don't simply teach you how I would approach problems.
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