## If an integer n is to be chosen at random from the integers

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ghong14
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### If an integer n is to be chosen at random from the integers

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

No sure how to really start with this. I know that for the denominator there has to be 96 possibilities and I can find all of the multiples of 8 between 8 and 92 but then how do I account for the odd numbers that can add up to a multiple of 8 as well. Thanks.
jnelson0612
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### Re: If an integer n is to be chosen at random from the integers

ghong14 wrote:

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

No sure how to really start with this. I know that for the denominator there has to be 96 possibilities and I can find all of the multiples of 8 between 8 and 92 but then how do I account for the odd numbers that can add up to a multiple of 8 as well. Thanks.

My first thought is to start writing out some possibilities and see what happens. Let's also keep in mind that all we need are three prime factors of 2 for the product to be divisible by 8. Let's try some various n's:

n=1 1 * 2 * 3 NO (not divisible by 8, because I don't have three prime factors of 2)
n=2 2 * 3 * 4 = YES
n=3 3 * 4 * 5= NO
n=4 4 * 5 * 6 = YES
n=5 5 * 6 * 7 = NO
n=6 6 * 7 * 8 = YES
n=7 7 * 8 * 9 = YES
n=8 YES (this one is obvious)
n=9 9 * 10 * 11 NO
n=10 10 * 11 * 12 YES

Okay, I think that we're seeing some things here:
1) every even n will create a product divisible by 8. Between the n and the n+2, we will have two even numbers, one of which will be divisible by 4. Thus, the two numbers are bringing at least three prime factors of 2.

Thus, every even n will work. Out of 96 numbers, 48 are even.

2) The only time an odd n is going to work is if n+1 is itself divisible by 8. Thus, the n's that work are:
n=7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95

I count them out and see that 12 work.

I then add 48 + 12 = 60. 60/96 n's will create a product divisible by 8, or 5/8.
Jamie Nelson
ManhattanGMAT Instructor
ghong14
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### Re: If an integer n is to be chosen at random from the integers

Brilliant explanation! Can't believe you are manning the forum on a FEDERAL Holiday!!! Thank you!
jnelson0612
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Joined: Fri Feb 05, 2010 10:57 am

### Re: If an integer n is to be chosen at random from the integers

ghong14 wrote:Brilliant explanation! Can't believe you are manning the forum on a FEDERAL Holiday!!! Thank you!

Haha, thanks so much, my pleasure! :-)
Jamie Nelson
ManhattanGMAT Instructor
PFWinkler
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### Re: If an integer n is to be chosen at random from the integers

Brilliant response. Thank you!
RonPurewal
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### Re: If an integer n is to be chosen at random from the integers

PFWinkler wrote:Brilliant response. Thank you!

did you mean to post this on the current thread?

doesn't seem like it (this discussion has been dormant for 3 years, and there is no "response" to anything that you've posted) ... but, figured I'd ask
PFWinkler
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### Re: If an integer n is to be chosen at random from the integers

Yes actually...Was just recently reviewing this problem. Thanks!
RonPurewal
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### Re: If an integer n is to be chosen at random from the integers

great, you're welcome.
ajaym8
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### Re: If an integer n is to be chosen at random from the integers

Umm..ok. I have one thing generalized in such questions.

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?

I check what's the divisor. From here I get to know ho many numbers should be in 1 'set' of numbers. So here I have to take 8 numbers in 1 set.
Now I just put n=1 to n=8 in the expression n(n + 1)(n + 2).
Exactly 5 numbers are divisible by 8 in such set of 8 numbers and I know that this trend will continue.

So probability = 5/8.

Is my method correct or is there some flaw in this ?

jnelson0612 wrote:
ghong14 wrote:

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

No sure how to really start with this. I know that for the denominator there has to be 96 possibilities and I can find all of the multiples of 8 between 8 and 92 but then how do I account for the odd numbers that can add up to a multiple of 8 as well. Thanks.

My first thought is to start writing out some possibilities and see what happens. Let's also keep in mind that all we need are three prime factors of 2 for the product to be divisible by 8. Let's try some various n's:

n=1 1 * 2 * 3 NO (not divisible by 8, because I don't have three prime factors of 2)
n=2 2 * 3 * 4 = YES
n=3 3 * 4 * 5= NO
n=4 4 * 5 * 6 = YES
n=5 5 * 6 * 7 = NO
n=6 6 * 7 * 8 = YES
n=7 7 * 8 * 9 = YES
n=8 YES (this one is obvious)
n=9 9 * 10 * 11 NO
n=10 10 * 11 * 12 YES

Okay, I think that we're seeing some things here:
1) every even n will create a product divisible by 8. Between the n and the n+2, we will have two even numbers, one of which will be divisible by 4. Thus, the two numbers are bringing at least three prime factors of 2.

Thus, every even n will work. Out of 96 numbers, 48 are even.

2) The only time an odd n is going to work is if n+1 is itself divisible by 8. Thus, the n's that work are:
n=7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95

I count them out and see that 12 work.

I then add 48 + 12 = 60. 60/96 n's will create a product divisible by 8, or 5/8.
RonPurewal
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### Re: If an integer n is to be chosen at random from the integers

that's correct, it's good enough just to check the numbers 1 through 8 (...or any other set of 8 consecutive integers).

you also need to check the given range of numbers, to see whether you have a whole number of complete "cycles" (without any numbers "left over" beyond these).
if the numbers are 1 through 96, then, yes, that's twelve complete groups of 8 consecutive integers.
(if you DIDN'T have a whole number of complete groups of 8, then you'd have to carefully account for the "leftover" numbers in your calculations.)