GMAT Forum

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- ghong14
- Course Students
**Posts:**105**Joined:**Wed Dec 31, 1969 8:00 pm

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
- ManhattanGMAT Staff
**Posts:**2664**Joined:**Fri Feb 05, 2010 10:57 am

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

ManhattanGMAT Instructor

- ghong14
- Course Students
**Posts:**105**Joined:**Wed Dec 31, 1969 8:00 pm

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

- jnelson0612
- ManhattanGMAT Staff
**Posts:**2664**Joined:**Fri Feb 05, 2010 10:57 am

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

ManhattanGMAT Instructor

- PFWinkler
- Course Students
**Posts:**32**Joined:**Wed Aug 29, 2012 3:11 am

Brilliant response. Thank you!

- RonPurewal
- Students
**Posts:**19747**Joined:**Tue Aug 14, 2007 8:23 am

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
- Course Students
**Posts:**32**Joined:**Wed Aug 29, 2012 3:11 am

Yes actually...Was just recently reviewing this problem. Thanks!

- RonPurewal
- Students
**Posts:**19747**Joined:**Tue Aug 14, 2007 8:23 am

great, you're welcome.

- ajaym8
- Students
**Posts:**24**Joined:**Tue Apr 12, 2016 7:32 pm

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 ?

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
- Students
**Posts:**19747**Joined:**Tue Aug 14, 2007 8:23 am

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.)

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.)

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