Probability

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dvalenz: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Wed Jul 20, 2011 11:40 am

by dvalenz » Fri Mar 29, 2013 9:37 am

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?

Thanks

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Source: Beat The GMAT — Problem Solving | Fri Mar 29, 2013 9:37 am

srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Fri Mar 29, 2013 9:47 am

there will be 2 cases in which n * n+1 * n+2 will be divisible by 8
CAse 1: N+1 is divisible by 8. How many factors of 8 are there from 1-96
that will be: 96/8 = 12 nos {8,16,24 .... 88,96}

Case 2:N = even because if n = even, n+2 will also be even
any 2 even numbers when multiplied will contain a 8 between them. eg 2 * 4 will contain 2^3
so essentially we need to know how many enen numbers are there between 1-96 and that will be 48 Nos

Total of such selections of N = 60 Nos
Total integers between 1-96 = 96 nos
hence prob. = 60 / 96 = 5/8
hence D

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Anju@Gurome: GMAT Instructor; Posts: 511; Joined: Wed Aug 11, 2010 9:47 amFollowed by: 86 members; Location: Delhi, India; Thanked: 344 times

by Anju@Gurome » Fri Mar 29, 2013 10:34 am

Note that n, (n + 1), and (n + 2) are 3 consecutive integers.
If the product of three consecutive integers are divisible by 8, then the following two scenarios are possible...

1. n is even --> (n + 1) is odd and (n + 2) is even --> Either n or (n + 2) will be divisible by 4
2. (n + 1) is divisible by 8 --> n and (n + 2) are odd

Now, the above two scenarios are mutually exclusive, i.e. no three consecutive integers can be part of both the scenarios. So, there are no repetitions and we can easily calculate number of triplets in each case and add them to get the total number of triplets.

Case 1: n is even
Number of even integers between 1 ans 96, inclusive = {(96 - 2)/2} + 1 = 48

Case 2: (n + 1) is divisible by 8
n + 1 = 8a, where a â‰¥ 1
n = 8a - 1
8a - 1 â‰¤ 96
8a â‰¤ 97
a â‰¤ 12.1 implies 12 integers.

Total number of n = (48 + 12) = 60
Therefore, probability that n(n + 1)(n + 2) will be divisible by 8 = 60/96 = 5/8

The correct answer is D.

Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Mar 29, 2013 1:09 pm

dvalenz 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?

Thanks

n(n+1)(n+2) = the product of 3 consecutive integers.

For many test-takers, the most efficient approach will be to WRITE IT OUT and LOOK FOR A PATTERN.

1*2*3
2*3*4
3*4*5
4*5*6
5*6*7
6*7*8
7*8*9
8*9*10

9*10*11
10*11*12
11*12*13
12*13*14
13*14*15
14*15*16
15*16*17
16*17*18

Each of the products in red is a multiple of 8.
The two examples above imply the following:
Of every 8 products, exactly 5 will be a multiple of 8.

Thus, the probability that n(n+1)(n+2) will be multiple of 8 = 5/8.

For a similar problem -- along with an alternate approach -- check my 2 posts here:

https://www.beatthegmat.com/probability-t116280.html

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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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