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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

30 Mar 2011, 02:37

1

This post received KUDOS

chloeholding wrote:

Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!

PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

A. 25% B 50% C 62.5% D. 72.5% E. 75%

N=n*(n+1)*(n+2)

Sol:

Let f(n) = n*(n+1)*(n+2)

The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.

f(2); the function becomes 2*3*4 will be divisible by 8 because 2 is even. f(52), the function becomes 52*53*54 will be divisible by 8 because 52 is even f(61), the function becomes 61*62*63 will NOT be divisible by 8 because 61 is odd

However, there is one more case where the function will be divisible by 8 even when first number of the three consecutive number is odd; f(7): the function becomes 7*8*9; the middle term 8 is divisible by 8 and thus the entire function becomes divisible by 8. f(63): the function becomes 63*64*65; the middle term 64 is divisible by 8 and thus the entire function becomes divisible by 8.

Probability = favorable/total

How many functions are there; f(1) = 1*2*3 f(2) = 2*3*4 f(3) = 3*4*5 f(4) = 4*5*6 . . . f(95) = 95*96*97 f(96) = 96*97*98

Total = 96

To find the favorable cases; we need to find the count of all even numbers from 1 to 96 and the count of all numbers that are divisible by 8.

Count of even numbers = (96-2)/2+1=47+1=48 Total numbers divisible by 8 = (96-8)/8+1 = 12

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

30 Mar 2011, 23:34

When n+1 is multiple of 8 it's just 96/8= 12 cases when f(n) = product of 3 consecutive integers is div by 8 Then when n is even we have 96/2=48 cases when f(n) is div by 8 Probability = (48+12)/96= 60/96 = 62.5%

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

31 Mar 2011, 21:44

PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

Total number of multiples of 8 between 1 to 96, inclusive = 12 sum of even numbers = [(first even+last even) / 2] - 1 = 48

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

03 Apr 2011, 06:34

Hello Fluke,

I understand your question. However, I have a small doubt. In this case when we are counting the even numbers, if they have a lot of multiples of 8 already included.

In that case, aren't we over counting the possibilities ?

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

03 Apr 2011, 07:00

pesfunk wrote:

Hello Fluke,

I understand your question. However, I have a small doubt. In this case when we are counting the even numbers, if they have a lot of multiples of 8 already included.

In that case, aren't we over counting the possibilities ?

Regards - Manish

Not Really!!!

f(n) = n*(n+1)*(n+2)

Now, the two cases I have mentioned above are mutually exclusive. f(6) = 6*7*8. Divisible by 8 because 6 is even i.e. n is even f(7) = 7*8*9. Divisible by 8 because 7 is odd and the middle term is a multiple of 8 i.e. (n+1) is multiple of 8. f(8) = 8*9*10. Divisible by 8 because 8 is even i.e. n=even

f(6) and f(8) will be counted in the even numbers case. f(7) will be counted in the case where we are finding the (n+1) or the middle number divisibility by 8.

Let's consider a smaller set. If n is an integer from 1 to 10 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

Here there are total 10 cases. n(1) not divisible by 8 because n=1(odd) and the middle number i.e. (n+1)=2 is not divisible by 8. n(2) is divisible by 8 because n=even=2 n(3) not divisible by 8 because n=3(odd) and the middle number i.e. (n+1)=4 is not divisible by 8. n(4) is divisible by 8 because n=even=4 n(5) not divisible by 8 because n=5(odd) and the middle number i.e. (n+1)=6 is not divisible by 8. n(6) is divisible by 8 because n=even=6 n(7) is divisible by 8 because n=odd=7 and (n+1)=8 is divisible by 8 n(8) is divisible by 8 because n=even=8 n(9) not divisible by 8 because n=9(odd) but the middle number i.e. (n+1)=10 is not divisible by 8. n(10) is divisible by 8 because n=even=10

How many cases are divisible by 8: n(2) - counted for the n=even case n(4) - counted for the n=even case n(6) - counted for the n=even case n(7) - counted for the n+1 divisible by 8 case n(8) - counted for the n=even case n(10) - counted for the n=even case Count=6 out of total 10 possibilities. Do you see any repetitions here? No, right.

With formula, if we just find the number of even numbers from 1 to 10 and numbers divisible by 8. We should have our result.

Even numbers from 1 to 10 = ((10-2)/2)+1 = 5 Numbers from 1 to 10 that are divisible by 8= (8-8)/8+1=0+1=1 Total=5+1=6
_________________

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

03 Apr 2011, 07:29

This explains pretty well!!

Many thanks Fluke for your VERY PROMPT response

fluke wrote:

pesfunk wrote:

Hello Fluke,

I understand your question. However, I have a small doubt. In this case when we are counting the even numbers, if they have a lot of multiples of 8 already included.

In that case, aren't we over counting the possibilities ?

Regards - Manish

Not Really!!!

f(n) = n*(n+1)*(n+2)

Now, the two cases I have mentioned above are mutually exclusive. f(6) = 6*7*8. Divisible by 8 because 6 is even i.e. n is even f(7) = 7*8*9. Divisible by 8 because 7 is odd and the middle term is a multiple of 8 i.e. (n+1) is multiple of 8. f(8) = 8*9*10. Divisible by 8 because 8 is even i.e. n=even

f(6) and f(8) will be counted in the even numbers case. f(7) will be counted in the case where we are finding the (n+1) or the middle number divisibility by 8.

Let's consider a smaller set. If n is an integer from 1 to 10 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

Here there are total 10 cases. n(1) not divisible by 8 because n=1(odd) and the middle number i.e. (n+1)=2 is not divisible by 8. n(2) is divisible by 8 because n=even=2 n(3) not divisible by 8 because n=3(odd) and the middle number i.e. (n+1)=4 is not divisible by 8. n(4) is divisible by 8 because n=even=4 n(5) not divisible by 8 because n=5(odd) and the middle number i.e. (n+1)=6 is not divisible by 8. n(6) is divisible by 8 because n=even=6 n(7) is divisible by 8 because n=odd=7 and (n+1)=8 is divisible by 8 n(8) is divisible by 8 because n=even=8 n(9) not divisible by 8 because n=9(odd) but the middle number i.e. (n+1)=10 is not divisible by 8. n(10) is divisible by 8 because n=even=10

How many cases are divisible by 8: n(2) - counted for the n=even case n(4) - counted for the n=even case n(6) - counted for the n=even case n(7) - counted for the n+1 divisible by 8 case n(8) - counted for the n=even case n(10) - counted for the n=even case Count=6 out of total 10 possibilities. Do you see any repetitions here? No, right.

With formula, if we just find the number of even numbers from 1 to 10 and numbers divisible by 8. We should have our result.

Even numbers from 1 to 10 = ((10-2)/2)+1 = 5 Numbers from 1 to 10 that are divisible by 8= (8-8)/8+1=0+1=1 Total=5+1=6

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

03 Apr 2011, 14:48

plb wrote:

fluke wrote:

The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even. Yes, because we will have product of two consecutive even numbers included in the expression.

Is this something that you derived, or is this a well-known property of 8?

It is a well known property. The product of two consecutive even numbers will always be divisible by 8.

Here's why; 8's prime factors = 2*2*2 2's prime factors = 2 4's prime factors = 2*2 2*4 prime factors = 2*2*2

Even if we don't take specifically 2 or 4. For any two consecutive even numbers; 1 number will be divisible by 2 and another divisible by 4.

Any two consecutive even numbers will definitely have "2*2*2" as its factors. And 8 also has "2*2*2" as its factor. Thus, the product of two consecutive even integers will always be divisible by 8.
_________________

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

04 Apr 2011, 07:08

Hi all, thanks so much for your replies, has been very helpful. I have a new problem though...

Question: For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100)+1, then p is a) between 2 and 10 b) between 10 and 20 c) between 20 and 30 d) between 30 and 40 e) greater than 40

Any replies would be greatly appreciated - thanks!

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

04 Apr 2011, 07:28

chloeholding wrote:

Hi all, thanks so much for your replies, has been very helpful. I have a new problem though...

Question: For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100)+1, then p is a) between 2 and 10 b) between 10 and 20 c) between 20 and 30 d) between 30 and 40 e) greater than 40

Any replies would be greatly appreciated - thanks!

chloeholding: Please post only one question per thread. The discussion for a new question should be opened in a separate thread.

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

05 Apr 2011, 07:43

fluke wrote:

chloeholding wrote:

Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!

PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

A. 25% B 50% C 62.5% D. 72.5% E. 75%

N=n*(n+1)*(n+2)

Sol:

Let f(n) = n*(n+1)*(n+2)

The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.

f(2); the function becomes 2*3*4 will be divisible by 8 because 2 is even. f(52), the function becomes 52*53*54 will be divisible by 8 because 52 is even f(61), the function becomes 61*62*63 will NOT be divisible by 8 because 61 is odd

However, there is one more case where the function will be divisible by 8 even when first number of the three consecutive number is odd; f(7): the function becomes 7*8*9; the middle term 8 is divisible by 8 and thus the entire function becomes divisible by 8. f(63): the function becomes 63*64*65; the middle term 64 is divisible by 8 and thus the entire function becomes divisible by 8.

Probability = favorable/total

How many functions are there; f(1) = 1*2*3 f(2) = 2*3*4 f(3) = 3*4*5 f(4) = 4*5*6 . . . f(95) = 95*96*97 f(96) = 96*97*98

Total = 96

To find the favorable cases; we need to find the count of all even numbers from 1 to 96 and the count of all numbers that are divisible by 8.

Count of even numbers = (96-2)/2+1=47+1=48 Total numbers divisible by 8 = (96-8)/8+1 = 12

first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12 so for each value there will be 3 pairs I.e. 12 x 3 =36 similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)

i added up 36+24=60 and hence the required probability = 60/96=.625

please correct me if i am wrong
_________________

WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

05 Apr 2011, 08:35

Warlock007 wrote:

Dear fluke I did it by simple method as below

first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12 so for each value there will be 3 pairs I.e. 12 x 3 =36 similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)

i added up 36+24=60 and hence the required probability = 60/96=.625

please correct me if i am wrong

You got the answer right but there seems some discrepancy in the counting.

When you counted all numbers divisible by 8 and then multiplied by 3; that would include sets such as

Below; 8 is divisible by 8 6*7*8 7*8*9 8*9*10

Below; 16 is divisible by 8 14*15*16 15*16*17 16*17*18

Below; 80 is divisible by 8 78*79*80 79*80*81 80*81*82

Sets where n=4 or a multiple of 4, such as 4*5*6 8*9*10 12*13*14 16*17*18

The colored sets are counted twice each.

At the same time, the following sets are never counted: 2*3*4 10*11*12

Luckily, the number of sets you missed compensated for the number of sets you double counted and got the correct result.
_________________

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

05 Apr 2011, 08:38

fluke wrote:

Warlock007 wrote:

Dear fluke I did it by simple method as below

first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12 so for each value there will be 3 pairs I.e. 12 x 3 =36 similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)

i added up 36+24=60 and hence the required probability = 60/96=.625

please correct me if i am wrong

You got the answer right but there seems some discrepancy in the counting.

When you counted all numbers divisible by 8 and then multiplied by 3; that would include sets such as

Below; 8 is divisible by 8 6*7*8 7*8*9 8*9*10

Below; 16 is divisible by 8 14*15*16 15*16*17 16*17*18

Below; 80 is divisible by 8 78*79*80 79*80*81 80*81*82

Sets where n=4 or a multiple of 4, such as 4*5*6 8*9*10 12*13*14 16*17*18

The colored sets are counted twice each.

At the same time, the following sets are never counted: 2*3*4 10*11*12

Luckily, the number of sets you missed compensated for the number of sets you double counted and got the correct result.

hey fluke you are right man....Thanks for your help
_________________

WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.

Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

Show Tags

08 Oct 2015, 02:00

chloeholding wrote:

Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!

PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

A. 25% B 50% C 62.5% D. 72.5% E. 75%

N=n*(n+1)*(n+2)

either n is divisible by 8 so 12 factors of 8 in the list, 12/96 = 1/8

or n is even which means 48 even numbers in list 48 /96 = 1/2

Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...