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
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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

The number of multiples of 3 from 1 to 99, inclusive, is:

(99 - 3)/3 + 1 = 33

We also should see that every number that is 1 less than a multiple of 3 will also allow n(n+1) to be divisible by 3. For example, if n = 2, then n + 1 = 3, which is divisible by 3. Since there are 33 multiples of 3, there are 33 numbers that are one less than a multiple of 3. Thus, the probability that n(n+1) will be divisible by 3 is 66/99 = 2/3.

Answer: D