Ask Questions, Discuss Approaches & No Clutter

If an integer n is to be chosen at random from the integers 1 to 96 inclusive

Solved 1 Answers 822 Views
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/
Official Answer

1 Best Answer

Best answer
n(n + 1)(n + 2) is divisible by 8 in two cases:
A. n=even, in this case n+2=even too and as n and n+2 are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. n+1 is itself divisible by 8;
(Notice that these two sets have no overlaps, as when n and n+2 are even then n+1 is odd and when n+1 is divisible by 8 (so even) then n and n+2 are odd.)
Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.
Answer: D.
answered Sep 2, 2014 by Partner (696 points)
selected Sep 2, 2014 by
Student Speaks

Jyotsna Mehta

Indian School of Business, Co'18

I am very thankful to Gmatxchange and would definitely recommend Gmatxchange mentors to all those students applying for MBA admissions

Your Tentative GMAT Score
Estimate your GMAT score.
The scorer helps you to determine your current level within a few questions you solve.Read More
Please log in or register to use the Scoring Feature
Confused about your profile & colleges, Get FREE profile evaluation today from over 10 consultants.
Follow us and get quick prep updates on Facebook