GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Sep 2019, 19:46

### 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

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 19 Nov 2011
Posts: 7
If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

Updated on: 28 Jan 2012, 05:12
29
2
230
00:00

Difficulty:

95% (hard)

Question Stats:

38% (02:17) correct 62% (02:14) wrong based on 1563 sessions

### HideShow timer Statistics

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

Originally posted by RadhaKrishnan on 28 Jan 2012, 04:03.
Last edited by Bunuel on 28 Jan 2012, 05:12, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 57996

### Show Tags

28 Jan 2012, 05:12
76
95
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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.
_________________
Senior Manager
Joined: 23 Oct 2010
Posts: 332
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38

### Show Tags

29 Jan 2012, 01:17
49
19
Bunuel 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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

from here u can also think this way (though it is a lil bit similar)
find how many numbers are divisible by 2 or 8
96/2=48
96/8=12
(48+12)/96=60/96=5/8
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth
##### General Discussion
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1011
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

18 Feb 2012, 08:03
28
11
+1 D

Other way is analyzing if there is a patron:

1) If n is an even number:
n:2, then 2*3*4 = 24 (divisible by 8)
n:4, then 4*5*6 = 120 (divisible by 8)
n:6, then 6*7*8= again divisible by 8
We have a patron.
So, we have 48 even possible values.

2) If n is an odd number:
This only can take place when n+1 is multiple of 8.
So, we have 12 possible values.

Then, $$\frac{(48 + 12)}{96} = \frac{5}{8}$$

D
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings
SVP
Joined: 24 Jul 2011
Posts: 1885
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740

### Show Tags

06 May 2012, 02:22
12
8
Any integer n(n+1)(n+2) will be divisible by 8 if n is a multiple of 2. This gives us 48 numbers between 1 and 96.

Additionally, all those numbers for which (n+1) is a multiple of 8 are also divisible by 8. This gives us a further 12 numbers. These numbers are all distinct from the first set because the first set had only even numbers and this set has only odd numbers.

Therefore probability = (48+12)/96 = 60/96 = 5/8

Option (D)
_________________

Awesome Work | Honest Advise | Outstanding Results

Reach Out, Lets chat!
Email: info at gyanone dot com | +91 98998 31738 | Skype: gyanone.services
Manager
Joined: 11 Sep 2013
Posts: 137
Concentration: Finance, Finance
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

17 Aug 2014, 00:24
12
1
goodyear2013 wrote:
If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?

A. 1/4
B. 1/2
C. 5/8
D. 3/4
E. 7/8

In such problem always follow a simple rule.
The divisor is 8. So take number from 1 to 8 as example
If n=1, n(n+1)(n+2)= Not divisible by 8
If n=2, n(n+1)(n+2)= yes
If n=3, n(n+1)(n+2)=No. Do upto 8

So, 5 out of 8 are divisible. Hence, 5/8 is the answer.

This rule is applicable if last number(96 in this case) is divisible by 8.
If you had to pick from 1-98, still you can apply the rule but be careful. But there is one additional step.
Hopefully you can find answer in less than 60 seconds
Senior Manager
Joined: 04 Jul 2014
Posts: 294
Location: India
GMAT 1: 640 Q47 V31
GMAT 2: 640 Q44 V34
GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

22 Nov 2014, 02:00
8
3
Here is the dumb way of doing this question, if the smart ones don't strike your head during the test

Since we are looking at divisibility by 8, we consider n to be between 1 and 8. We can multiply the result by 12 - since 96/8 = 12.

Attachment:

Capture.PNG [ 6.47 KiB | Viewed 84080 times ]

_________________
Cheers!!

JA
If you like my post, let me know. Give me a kudos!
Math Expert
Joined: 02 Sep 2009
Posts: 57996
If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

25 Jul 2014, 03:54
4
3
Senior Manager
Joined: 02 Mar 2012
Posts: 281
Schools: Schulich '16
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

17 May 2013, 13:07
3
there are total 48 numbers in the form n*n+1*n+2 starting from 2,4,6,8,10.....96 for which it is divisible by 8.

Additionally,there are 12 cases if there is 8 or a multiple of 8 that also divides the form n*n+1*n+2 which starts from 7,15,23,31....95.

The above cases both are non-overlapping, so we can add them.

adding the above two cases 48+12=60
ans=60/96=5/8 which is D.

--harry
Do press kudos if u like my post
Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 130
Location: India
WE: Information Technology (Investment Banking)

### Show Tags

30 Jan 2012, 09:10
2
1
Bunuel 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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.

Bunnel - Thanks for the explanation. I did not take the condition when n+1 is also divisible by 8
Math Expert
Joined: 02 Sep 2009
Posts: 57996

### Show Tags

08 May 2012, 01:05
2
Bunuel wrote:
anjolaolu wrote:
(PLS CAN SOMEONE KINDLY EXPLAIN WHY THE ANSWER OF THE QUESTION BELOW IS D, THANK YOU)

If an integer n is to be chosen at random from 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

Hi,
please correct where i am going wrong with my apporach.

between 1 and 96 inclusive, there are 12 multiples of 8 i.e. 8,16,24,32,40,48,54,64,72,80,88,96. so, if n is any of the 12 numbers, then n.(n+1)(n+2) is divisible by 8.

now, based on the above listed, the numbers for n+1 can be 7,15,23,31,39,47,53,63,71,79,87,95 and similary for
n+2 = 6,14,22,30,38,46,52,62,70,78,86,96.

so, a total of 36, hence 36/98 = 3/8 which is incorrect.

thanks
jay

There are more cases for n(n+1)(n+2) to be divisible by 8. Please read the solution above.

n(n+1)(n+2) is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8.
_________________
CEO
Joined: 20 Mar 2014
Posts: 2613
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

12 Jul 2015, 05:40
2
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

Alternate solution:

Look at the first few sets of 3s:
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9
8,9,10

We see that out of the above 8 sets, favorable cases are 5 (in red). Thus the probability is 5/8. This will repeat till we have 88,89,90 making it 9 total patterns.

Now consider after 88,89,90, we get

89,90,91
90,91,92
91,92,93
92,93,94
93,94,95
94,95,96
95,96,97
96,97,98

So we have another 5 favorable out of the remaining 8 sets. Thus we have final 5/8 as the probability.

Thus the final probability = {(5/8)*9+(5/8)*1} / (9+1) = 5/8
Intern
Joined: 03 Apr 2012
Posts: 23

### Show Tags

07 May 2012, 13:07
1
Bunuel wrote:
anjolaolu wrote:
(PLS CAN SOMEONE KINDLY EXPLAIN WHY THE ANSWER OF THE QUESTION BELOW IS D, THANK YOU)

If an integer n is to be chosen at random from 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

Hi,
please correct where i am going wrong with my apporach.

between 1 and 96 inclusive, there are 12 multiples of 8 i.e. 8,16,24,32,40,48,54,64,72,80,88,96. so, if n is any of the 12 numbers, then n.(n+1)(n+2) is divisible by 8.

now, based on the above listed, the numbers for n+1 can be 7,15,23,31,39,47,53,63,71,79,87,95 and similary for
n+2 = 6,14,22,30,38,46,52,62,70,78,86,96.

so, a total of 36, hence 36/98 = 3/8 which is incorrect.

thanks
jay
Manager
Joined: 30 May 2013
Posts: 152
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

### Show Tags

19 Aug 2013, 01:39
1
Bunuel 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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.

Hi,

Can we use this formula for this problem
# of multiples of x in the\ range = frac{Last multiple of x in the range - First multiple of x in the range}{x}+1,

Which give value of = 12

total outcome is 96,

Please explain this with the above formula.

Regards,
Rrsnathan.
Intern
Joined: 13 Dec 2011
Posts: 47
GPA: 4
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

04 Sep 2014, 13:42
1
Bunuel: Please suggest where I went wrong;

Although I understand your solution, I have a doubt on my approach.

I multiplied n(n+1)(n+2) and I got n3+3n2+2n ... raised to the power;

So what I thought was...I need to find the numbers for which the quotient is zero for the above expression....since the last number of the above expression is 2n, it just needs a minimum 4 to be divisible by 8. Moreover a minimum n=4 will make n3 and 3n2 easily divisible by 8; Therefore, I thought that the above expression will be true for all the multiples of 4...there are 24 multiples of 4 from 1 to 96 inclusive, hence my answer comes out to be 24/96=1/4;

I will be really thankful to you if you can tell me where did I go wrong!!

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 57996
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

05 Sep 2014, 06:05
1
execnitinsharma wrote:
Bunuel: Please suggest where I went wrong;

Although I understand your solution, I have a doubt on my approach.

I multiplied n(n+1)(n+2) and I got n3+3n2+2n ... raised to the power;

So what I thought was...I need to find the numbers for which the quotient is zero for the above expression....since the last number of the above expression is 2n, it just needs a minimum 4 to be divisible by 8. Moreover a minimum n=4 will make n3 and 3n2 easily divisible by 8; Therefore, I thought that the above expression will be true for all the multiples of 4...there are 24 multiples of 4 from 1 to 96 inclusive, hence my answer comes out to be 24/96=1/4;

I will be really thankful to you if you can tell me where did I go wrong!!

Thanks.

You are missing cases there. Won't n^3+3n^2+2n when n is any even number? Also it can be divisible by 8, even if individual terms are not but the sum is. For example check for n =7.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 57996
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

16 Jul 2016, 02:41
1
megha_2709 wrote:
Bunuel 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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.

Hi,

Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain.
Sorry if this sounds too basic.

Regards
Megha

In EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8
_________________
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 7720
Location: United States (CA)
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

16 Nov 2016, 08:55
1
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

We are given that an integer n is to be chosen at random from the integers 1 to 96 inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8.

We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers.

Case 1: n is even. Any time that n is even, n + 2 will also be even. Moreover, either n or n + 2 will be divisible by 4, and thus n(n + 1)(n + 2) will contain three factors of 2 and will be divisible by 8.

Since there are 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96 inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8.

Case 2: n is odd. If n is odd, then n(n + 1)(n + 2) still can be divisible by 8 if the factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96 inclusive.

Number of multiples of 8 = (96 - 8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8.

In total, there are 48 + 12 = 60 outcomes in which n(n + 1)(n + 2) will be divisible by 8.

Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 60/96 = 10/16 = 5/8.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Joined: 25 Dec 2016
Posts: 17
Location: United States (GA)
Concentration: Healthcare, Entrepreneurship
GMAT 1: 770 Q51 V42
GPA: 3.64
WE: Medicine and Health (Health Care)
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

07 Feb 2017, 07:48
1
My approach was similar to Bunuel's and AdmitJA's, but I wanted to offer it because it was simple and quick. I was able to do this one in just over a minute.

When I read the question, I paid attention to two things right away. The first was that $$(n)(n+1)(n+2)$$ is the product of three consecutive integers. Because $$8=2^3$$, I knew that I needed to find three 2s in the factorization of whatever 3 consecutive numbers I used. The second thing that I noticed is that 96 is divisible by 8. This indicated to me that whatever pattern I noticed in the first 8 numbers would be repeated 12 times through 96. Since I recognized that the pattern would be repeated, I knew that I only needed to look at the first 8 numbers.

I listed the numbers out: 1 2 3 4 5 6 7 8.

Since 8 has three 2s in its factorization, I knew that n=6,7,8 would all be divisible by 8. I saw that n=2 would work because 2 has one 2 and 4 has two 2s. Similarly, I saw that n=4 would work because 4 has two 2s and 6 has one 2. That gives us 5 options in the first 8 numbers. So the answer is 5/8.

As mentioned above, this ratio will be the same for every 8 numbers, so 5/8 will be true of 8x where x is any positive integer. A trickier version of this question would have been to make the number NOT divisible by 8. In that case, I think you should still find the pattern for every 8 numbers, but you'd also want to look at the "extra" numbers to figure out the fraction.

Take 53, for instance. You'd want to recognize that 8 goes into 53 six times with a remainder of 5. This means that you'd have the $$5/8$$ ratio for 6 sets of 8 but also an "incomplete" set of n = 1,2,3,4,5. In the first 48 numbers, you'd have 30 that would be divisible by 8. In the "incomplete" set you'd have 2 (since n=2 and n=4 are both divisible by 8). Thus you'd have 32/53 numbers divisible by 8.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9633
Location: Pune, India
Re: If an integer n is to be chosen at random from the integers  [#permalink]

### Show Tags

15 May 2018, 10:26
1
siddreal 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

VeritasPrepKarishma Pls show us your approach to this question. Though I understood the solutions mentioned in the thread, I am finding it difficult to believe that I might think of this approach during actual exam.

What I would do is pretty much what Bunuel has done in his solution.

Note that the moment I see n(n + 1)(n + 2), I think of divisibility in a bunch of consecutive numbers.
In 3 consecutive integers, if n is even, (n+2) is even too. If there are two consecutive even integers, one of them will be a multiple of 4.
So if n is even, (n+2) is even too and one of them is definitely a multiple of 4.
So n(n + 1)(n + 2) becomes divisible by 8 in each case that n is even.
From 1 to 96, half the cases have even n so this mean 48 cases.

Alternatively, when n is odd, n+1 is even. But then n+2 is odd too. So to be a multiple of 8, (n+1) will need to be a multiple of 8.
Hence this gives us another 12 cases (n+1 goes from 8 to 96).
Note that there will be no overlap in the two since here n is definitely odd.

Total we have 60 cases of the possible 96 which gives 60/96 = 5/8

For these properties of numbers, see
https://www.veritasprep.com/blog/2011/0 ... c-or-math/
https://www.veritasprep.com/blog/2011/0 ... h-part-ii/
_________________
Karishma
Veritas Prep GMAT Instructor

Re: If an integer n is to be chosen at random from the integers   [#permalink] 15 May 2018, 10:26

Go to page    1   2   3    Next  [ 43 posts ]

Display posts from previous: Sort by