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

 It is currently 20 Oct 2019, 22:14

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 1 to 96

Author Message
TAGS:

Hide Tags

GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4015
Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

Show Tags

20 Apr 2018, 13:32
Top Contributor
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

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Joined: 06 Mar 2017
Posts: 189
Concentration: Operations, General Management
Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

Show Tags

15 May 2018, 10:06
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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: If an integer n is to be chosen at random from the integers 1 to 96  [#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

CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

Show Tags

22 Oct 2018, 23:35
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

Please find the video solution here.

Attachments

File comment: www.GMATinsight.com

Screenshot 2018-10-23 at 11.30.09 AM.png [ 682.59 KiB | Viewed 651 times ]

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

Show Tags

13 Jan 2019, 06:36
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

$$?\,\,\, = \,\,\,{{{?_{\,{\rm{temp}}}}\,\,\,\,\left( {{\rm{favorable}}\,\,{\rm{remainders}}} \right)} \over {8\,\,\,\,\left( {{\rm{equiprobable}}\,\,{\rm{remainders}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{division}}\,\,{\rm{by}}\,\,{\rm{8}}} \right)}}$$

$$\left. \matrix{ n\,\, \in \,\,\left\{ {8M,8M + 2,8M + 4,8M + 6} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n\;\;{\rm{even}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {n + 1} \right)\left( {n + 2} \right)} \over 8} = {\mathop{\rm int}} \,\,\,\, \hfill \cr n\,\, \in \,\,\left\{ {8M + 1,8M + 3,8M + 5} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {n + 1} \right)\left( {n + 2} \right)} \over 8} \ne {\mathop{\rm int}} \hfill \cr n\,\, \in \,\,\left\{ {8M + 7} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {8M + 8} \right)\left( {n + 2} \right)} \over 8} = {\mathop{\rm int}} \hfill \cr} \right\}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{?_{\,{\rm{temp}}}} = 5$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Manager
Joined: 18 Jul 2015
Posts: 73
GMAT 1: 530 Q43 V20
WE: Analyst (Consumer Products)
Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

Show Tags

15 Aug 2019, 23:59
It took me some time to solve this one, but here's my take on the question.

1. When selecting an integer from 1 to 96, there will be 2 scenarios:
1.a Selecting an even integer in which case $$n(n+1)(n+2)$$ will be $$even * odd * even$$
1.b Selecting an odd integer in which case $$n(n+1)(n+2)$$ will be $$odd * even * odd$$

1.a:
Let's start with the smallest even integer that we can pick i.e. 2. In this case the expression will be $$2*3*4$$, which is divisible by 8. Next take 4 and the expression will be $$4*5*6$$ again divisible by 8. As you can see picking up any even will have three 2's in the expression and hence the expression will be divisible by 8. To conclude, picking any even will make the expression divisible by 8.

Hence, it now boils down to how many even integers can we pick from 1 to 96 $$\frac{(96-2)}{2} +1 = 48$$

1.b:
The two odd integers will not be divisible by 8, hence we need to focus on the middle even integer. Let's start with 1 and the expression will be $$1*2*3$$, clearly not divisible by 8. Next 3 and the expression will be $$3*4*5$$, clearly not divisible by 8. Next 5 and the expression will be $$5*6*7$$, clearly not divisible by 8. Next 7 and the expression will be $$7*8*9$$ and we got the first clear win. Next 9 and the expression will be $$9*10*11$$, clearly not divisible by 8. We can conclude that if the middle integer is a multiple of 8 then the expression will be divisible by 8.

Hence, it now boils down to how many odd integers can we pick from 1 to 96 such that the middle integer is a multiple of 8 or how many multiples of 8 can we pickup from 1 to 96$$=\frac{(96-8)}{8}+1 = 12$$

Combining 1.a and 1.b we get,

$$\frac{Favorable Outcomes}{Total Outcomes}$$=$$\frac{(48 + 12)}{96}$$=$$\frac{60}{96}$$=$$\frac{5}{8}$$ (Ans. D)

P.S The 96 in the denominator came from "If an integer n is to be chosen at random from the integers 1 to 96, inclusive" i.e Total Outcomes $$(96-1)+1$$
_________________
Cheers. Wishing Luck to Every GMAT Aspirant!
Re: If an integer n is to be chosen at random from the integers 1 to 96   [#permalink] 15 Aug 2019, 23:59

Go to page   Previous    1   2   [ 26 posts ]

Display posts from previous: Sort by