Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: integers and prob [#permalink]
19 Jan 2008, 22:01

marcodonzelli 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

1,2,3,...96.

n can be chosen in 96 ways. ( This is total number of samples )

n*(n+1)*(n+2)/8 implies we should find number of favorable samples in which we have

n*(n+1)*(n+2) = 8(min), 16, 32, ...96(max)

8 = 2^3 ( Thus total number of divisors of 8 will be (3+1)=4 ) which are

1, 2, 4, 8. From this we get 3 different possibilities;

2*(2+1)*(2+2) 4*(4+1)*(4+2) 8*(8+1)*(8+2)

But with 8 we can also have;

6*(6+1)*(6+2) 7*(7+1)*(7+2)

Similarly for 16 = 2^4 ( Thus total number of divisors of 8 will be (4+1)=5 ) which are

1, 2, 4, 8, 16. From this also we get 4 different possibilities;

2*(2+1)*(2+2) 4*(4+1)*(4+2) 8*(8+1)*(8+2) Two more associated with it as shown above. 16(16+1)*(16+2) Two more associated with it depending on the position of 16.

Two of them are already in the sequence, so 3 different.

and so on... with each multiple of 8, we will get 3 different possibilities.

Now, how many are the total number of number which are multiple of 8 from 8...96. (96-8/8)+1=12

Each of these 12 different multiples have 3 different possibilities in which 8 can be the divisor.

Thus total number of favorable outcomes = 12*3 + 2*(2+1)*(2+2) + 4*(4+1)*(4+2) = 36+1+1

Re: integers and prob [#permalink]
24 Jan 2008, 10:43

C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2 whats the OA pls?????

Re: integers and prob [#permalink]
24 Jan 2008, 10:49

Expert's post

Dellin wrote:

C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2

read above.

plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8 _________________

Re: integers and prob [#permalink]
20 Feb 2008, 00:46

walker wrote:

Dellin wrote:

C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2

read above.

plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8

walker can you help me with this approach?

96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?

Re: integers and prob [#permalink]
20 Feb 2008, 01:35

Expert's post

marcodonzelli wrote:

walker can you help me with this approach?

96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?

I think the approach is not correct. You missed the fact that we have 3 consecutive integers. For example, 4,5,6 are not multiplies of 8 but 4*5*6 is divisible by 8. _________________

Re: integers and prob [#permalink]
20 Feb 2008, 09:20

marcodonzelli 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

Total numbers 8*12 There are 12 numbers divisible by 8 -> 3*12 (if 8 is an example - (6,7,8), (7,8,9), (8,9,10)) and 12 numbers divisible by 4 but not divisible by 8 -> 2*12 (if 4 is an example (2,3,4) and (4,5,6))

Re: integers and prob [#permalink]
17 Mar 2008, 02:32

1

This post received KUDOS

marcodonzelli wrote:

StartupAddict wrote:

My 30 second answer:

if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.

then, we consider if n+1 is divislbe by 8, which is another 12 times

so 48 + 12 / 96 = 60/96 = 30/48 = 5/8

QED

that's a great job really. quick and easy. kudo for you

I didn't imagine it could be so easy, without any use of calculus and without trying with numbers, somewhat very difficult to do in 2 mins during the exam:

Re: integers and prob [#permalink]
28 Apr 2008, 00:09

OK, here is a long explanation for a problem that took me less than a minute.

First and most important, we know that for n(n+1)(n+2) to be divisible by 8, either n or n+1 or n+2 must be divisible by 8

Second we have to realize that we are dealing with consecutive integers.

We also realize that as we have consecutive integers and multiplication, every even integer n from 2 to 96 satisfy n(n+1)(n+2) to be divisible by 8, allowing us to quickly count all even integers from 2 to 96 to be 47, when we look when n or n+1 or n+2 is divisible by 8.

Realizing the above three in a snap, we simply check when n or n+1 or n+2 is divisible by 8:

1. n is divisible by 8 when n is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

2. n+1 is divisible by 8, when n+1 is 8,16,32,36,40,48, 56,64,72,80,88,96, but as we deal with consecutive integers, this means that in such case, n will be odd integer 7,31,35,39,47,55,63,71,79,87,95, therefore we can pick another 13 integers that make n(n+1)(n+2) divisible by eight. We have to add those 13 odds to our 47 evens.

3. n+2 is divisible by 8, when n+2 is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

we have 47+13 integers satisfying n(n+1)(n+2) is divisible by 8, thus the probability of picking one of these integers is simply P(E) = n of event occuring/total number, that is 60/96, which is equal to 5/8.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...