NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 09:37 It is currently 26 Apr 2024, 09:37
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

Tough and tricky: Probability of integr being divisible by 8

User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92945
Own Kudos [?]: 619197 [4]
Given Kudos: 81609
Send PM
CAT Tests Forum Quiz Mode
Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   11 Oct 2009, 18:33
4
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

47% (02:09) correct 53% (01:40) wrong based on 108 sessions
Hide Show timer Statistics
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

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

Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
hgp2k
Manager
Manager
Joined: 18 Aug 2009
Posts: 192
Own Kudos [?]: 773 [0]
Given Kudos: 13
 Q50  V35
Send PM
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   11 Oct 2009, 18:39
There could be 36 such integers. So probability according to me should be 36/96 = 37.5%.
Can you please recheck the answer choices?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92945
Own Kudos [?]: 619197 [0]
Given Kudos: 81609
Send PM
CAT Tests Forum Quiz Mode
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   11 Oct 2009, 18:54
Expert Reply
hgp2k wrote:
There could be 36 such integers. So probability according to me should be 36/96 = 37.5%.
Can you please recheck the answer choices?


Checked answered choices, then solved again myself, seems choices are correct. Can you please share your way of thinking?
User avatar
hgp2k
Manager
Manager
Joined: 18 Aug 2009
Posts: 192
Own Kudos [?]: 773 [0]
Given Kudos: 13
 Q50  V35
Send PM
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   11 Oct 2009, 19:08
Sure. Here is what I think, and I might be wrong (In fact most of the times I am wrong!):

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8.
Now, there are 96/8 = 12 integers from 1 to 96 which are divisible by 8.
For n*(n+1)*(n+2) to be divisible by 8: n, (n+1) or (n+2) must be divisible by 8. There could be 12*3 = 36 such integers.
So the probability for n*(n+1)*(n+2) being divisible by 8 = 36/96. = 3/8 = 37.5%

Please correct me if I am wrong.
avatar
bkuruvalli
Intern
Intern
Joined: 18 Aug 2005
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 0
Send PM
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   11 Oct 2009, 22:18
Hmm. isnt it C - 62.5


Here is my logic behind it..

The product is a combination of 3 numbers. There a total of 96 of these.

Any combination with two even numbers (starting with n=2) can be divided by 8. - 48 possible outcomes

Of the combinations with two even numbers the ones with multiples of 8 (eg: when n=7,15,23) are also divisible by 8. - 12 possible outcomes.

60/96 = 62.5%
User avatar
verbalfight
Intern
Intern
Joined: 12 Jul 2009
Posts: 30
Own Kudos [?]: 16 [0]
Given Kudos: 9
Send PM
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   11 Oct 2009, 22:39
I ll go with C. Whats the OA ???
avatar
rohitbhotica
Intern
Intern
Joined: 07 Oct 2009
Posts: 11
Own Kudos [?]: 16 [3]
Given Kudos: 0
Send PM
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   12 Oct 2009, 04:11
2
Kudos
1
Bookmarks
answer is C

If n is odd then only 12 choices - 7,15,23 etc....

if n is even then for all n it is divisible by 8
as n*(n+2) is divisible by 8 for all even n

Hence total numbers divisible by 8 = 48+12 = 60

Answer = 60/96*100% = 62.5%
User avatar
hgp2k
Manager
Manager
Joined: 18 Aug 2009
Posts: 192
Own Kudos [?]: 773 [0]
Given Kudos: 13
 Q50  V35
Send PM
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   12 Oct 2009, 05:32
hmmmm, I got it now. I was considering only the multiples of 8 and not all even numbers.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92945
Own Kudos [?]: 619197 [0]
Given Kudos: 81609
Send PM
CAT Tests Forum Quiz Mode
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   12 Oct 2009, 09:29
Expert Reply
Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html

You can continue discussions and see the solutions there.
User avatar
B
dimitri92
Manager
Manager
Joined: 15 Nov 2006
Affiliations: SPG
Posts: 232
Own Kudos [?]: 3138 [3]
Given Kudos: 34
Send PM
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink] New post   11 Feb 2011, 05:39
3
Kudos

this questions tests the concepts of consecutive integers.

three consecutive integers will ..
1. have an integer that is a multiple of 3 e.g. {2,3,4} or {3,4,5}
2. either have two odds or two even integers e.g. {2,3,4} or {3,4,5}
3. have two even integers if n (1st integer) is even. the product n*(n+1)*(n+2) must be a multiple of 8 because one even integer will be a multiple of 4. e.g. {2,4,5}, {4,5,6} or {14,15,16}
4. have two odd integers if n (1st integer) is odd. the product can be a multiple of 8 only if (n+1) is a multiple of 8 because the other two are odd. e.g. {7,8,9} or {23,24,25}. on the other hand, the product of {3,4,5} or {11,12,13} is not multiple of 8 because (n+1) is not multiple of 8.

we can use the above rules to calculate probability.

no. of cases where n is even = \(\frac{96}{2} = 48\)
no. of cases where n+1 is multiple of 8 = \(\frac{96}{8} = 12\)
total cases in which product is multiple of 8 = \(48+12 = 60\)

probability = \(\frac{60}{96}\) = 62.5%

Ans: C


Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]
11 Feb 2011, 05:39
Moderators:
Bunuel
Math Expert
92945 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions