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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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...
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 ≥ 1
n = 8a - 1
8a - 1 ≤ 96
8a ≤ 97
a ≤ 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.
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
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 ≥ 1
n = 8a - 1
8a - 1 ≤ 96
8a ≤ 97
a ≤ 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.
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n(n+1)(n+2) = the product of 3 consecutive integers.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
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.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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