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.

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:

Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to pre-think assumptions and solve the most challenging questions in less than 2 minutes.

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

RadhaKrishnan 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

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.

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

RadhaKrishnan 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.

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:

RadhaKrishnan 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

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

RadhaKrishnan 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

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

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\(=\frac{(96-8)}{8}+1 = 12\)

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 \((96-1)+1\)
_________________