AAPL wrote: ↑Tue Apr 14, 2020 3:30 am
GMAT Prep
An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3?
A. 1/9
B. 1/3
C. 1/2
D. 2/3
E. 5/6
OA
D
Let's analyze the pattern of n(n + 1).
1. @n = 1, we have n(n + 1) = 1*2: not divisible by 3
2. @n = 2, we have n(n + 1) = 2*3: divisible by 3
3. @n = 3, we have n(n + 1) = 3*4: divisible by 3
4. @n = 4, we have n(n + 1) = 4*5: not divisible by 3
5. @n = 5, we have n(n + 1) = 5*6: divisible by 3
6. @n = 6, we have n(n + 1) = 6*7: divisible by 3
You would observe that for every three values of n, one out of them makes n(n + 1) not divisible by 3 and two out of them make n(n + 1) divisible by 3.
Thus, out of 99 values of n, we have 2/3 values of n that make n(n + 1) divisible by 3.
The correct answer:
D
Hope this helps!
-Jay
_________________
Manhattan Review Test Prep
Locations:
Manhattan Review Bangalore |
Chennai |
GRE Prep Hyderabad |
Himayatnagar GRE Coaching | and many more...
Schedule your free consultation with an experienced GMAT Prep Advisor!
Click here.
