Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 May 2016, 20:39

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If n is an integer from 1 to 96 (inclusive), what is the probability

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 30 Mar 2011
Posts: 20
Followers: 0

Kudos [?]: 4 [0], given: 0

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

### Show Tags

30 Mar 2011, 02:09
00:00

Difficulty:

65% (hard)

Question Stats:

44% (01:56) correct 56% (01:53) wrong based on 43 sessions

### HideShow timer Statistics

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%

[Reveal] Spoiler:
N=n*(n+1)*(n+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!!
[Reveal] Spoiler: OA
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 154

Kudos [?]: 1436 [1] , given: 376

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

### Show Tags

30 Mar 2011, 02:37
1
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

$$Probability = \frac{48+12}{96} = \frac{60}{96}=0.625=62.5%$$

Ans: "C"

Please see this as well:
http://gmatclub.com/forum/tough-tricky-set-of-problms-85211-40.html#p799275
_________________
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 13759
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3279

Kudos [?]: 18878 [0], given: 4237

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

### Show Tags

30 Mar 2011, 11:43
Expert's post
chloeholding, please post your problems in either PS or DS subforums
_________________

Founder of GMAT Club

US News 2008 - 2017 Rankings progression - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

GMAT Club Premium Membership - big benefits and savings

SVP
Joined: 16 Nov 2010
Posts: 1673
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 34

Kudos [?]: 440 [0], given: 36

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

### Show Tags

30 Mar 2011, 19:53
@fluke

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;

3 * 4 * 5 is not divisible by 8, right, perhaps it's more accurate to say the odd number should be at least 7 ?

Also, just to find the number of integers "(n+1)" divisible by 8 from 1 - 96 in this case :

95 = 7 + (N-1)*8 (N is total such numbers, and it's an AP)

=> (N-1)*8 = 88

=> N = 12
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 920
Followers: 13

Kudos [?]: 292 [0], given: 123

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%

Posted from my mobile device
Intern
Joined: 11 Oct 2010
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

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

Probability = 48+12 / 96 = 60 / 96 = 62.5%
Intern
Joined: 06 Sep 2010
Posts: 45
Followers: 0

Kudos [?]: 4 [0], given: 0

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 ?

Regards - Manish
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 154

Kudos [?]: 1436 [0], given: 376

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)

n(1) = 1*2*3
n(2) = 2*3*4
n(3) = 3*4*5
n(4) = 4*5*6
n(5) = 5*6*7
n(6) = 6*7*8
n(7) = 7*8*9
n(8) = 8*9*10
n(9) = 9*10*11
n(10) = 10*11*12

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
_________________
Intern
Joined: 06 Sep 2010
Posts: 45
Followers: 0

Kudos [?]: 4 [0], given: 0

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)

n(1) = 1*2*3
n(2) = 2*3*4
n(3) = 3*4*5
n(4) = 4*5*6
n(5) = 5*6*7
n(6) = 6*7*8
n(7) = 7*8*9
n(8) = 8*9*10
n(9) = 9*10*11
n(10) = 10*11*12

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
Intern
Joined: 08 Dec 2010
Posts: 10
Followers: 0

Kudos [?]: 3 [0], given: 1

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

### Show Tags

03 Apr 2011, 14:37
fluke wrote:
The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.

Is this something that you derived, or is this a well-known property of 8?
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 154

Kudos [?]: 1436 [0], given: 376

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.
_________________
Intern
Joined: 30 Mar 2011
Posts: 20
Followers: 0

Kudos [?]: 4 [0], given: 0

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!
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 154

Kudos [?]: 1436 [0], given: 376

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.

Here's the link where you can discuss this new question:
http://gmatclub.com/forum/hard-question-function-and-factors-90863.html
_________________
Manager
Status: ==GMAT Ninja==
Joined: 08 Jan 2011
Posts: 247
Schools: ISB, IIMA ,SP Jain , XLRI
WE 1: Aditya Birla Group (sales)
WE 2: Saint Gobain Group (sales)
Followers: 5

Kudos [?]: 67 [0], given: 46

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

$$Probability = \frac{48+12}{96} = \frac{60}{96}=0.625=62.5%$$

Ans: "C"

Please see this as well:
http://gmatclub.com/forum/tough-tricky-set-of-problms-85211-40.html#p799275

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
_________________

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.

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 154

Kudos [?]: 1436 [0], given: 376

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.
_________________
Manager
Status: ==GMAT Ninja==
Joined: 08 Jan 2011
Posts: 247
Schools: ISB, IIMA ,SP Jain , XLRI
WE 1: Aditya Birla Group (sales)
WE 2: Saint Gobain Group (sales)
Followers: 5

Kudos [?]: 67 [0], given: 46

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.

Director
Joined: 01 Feb 2011
Posts: 758
Followers: 14

Kudos [?]: 93 [0], given: 42

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

### Show Tags

21 Apr 2011, 19:00
when n is even n (n+1)(n+2) is divisible by 8. there are 48 possible combinations for this.

n is odd and n+1 divisible by 8( i.e n = 8P-1) there are 12 possibilities that match this.

= 60/96 = 62.5

Intern
Joined: 02 Mar 2015
Posts: 32
Followers: 0

Kudos [?]: 3 [0], given: 8

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

so total is 5/8 = 62.5 %
Math Expert
Joined: 02 Sep 2009
Posts: 33000
Followers: 5752

Kudos [?]: 70489 [0], given: 9847

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

### Show Tags

08 Oct 2015, 07:31
Expert's post
chloeholding wrote:
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%

[Reveal] Spoiler:
N=n*(n+1)*(n+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!!

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
_________________
Re: If n is an integer from 1 to 96 (inclusive), what is the probability   [#permalink] 08 Oct 2015, 07:31
Similar topics Replies Last post
Similar
Topics:
6 If n is the product of the integers from 1 to 8, inclusive 10 09 Jul 2012, 03:31
5 If n is an integer from 1 to 96, what is the probability for 7 14 Nov 2011, 05:56
63 If n is the product of integers from 1 to 20 inclusive 21 14 Dec 2010, 18:31
If n is the product of the integers from 1 to 20 inclusive, 4 22 Nov 2009, 18:05
1 If n is the product of the integers from 1 to 8, inclusive, 4 22 Aug 2007, 11:22
Display posts from previous: Sort by

# If n is an integer from 1 to 96 (inclusive), what is the probability

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.