## An integer $$n$$ between $$1$$ and $$99,$$ inclusive, is to be chosen at random. What is the probability that $$n(n+1)$$

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### An integer $$n$$ between $$1$$ and $$99,$$ inclusive, is to be chosen at random. What is the probability that $$n(n+1)$$

by Vincen » Wed Mar 31, 2021 8:13 am

00:00

A

B

C

D

E

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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) $$\dfrac19$$

B) $$\dfrac13$$

C) $$\dfrac12$$

D) $$\dfrac23$$

E) $$\dfrac56$$

Answer: D

Source: GMAT Prep

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### Re: An integer $$n$$ between $$1$$ and $$99,$$ inclusive, is to be chosen at random. What is the probability that $$n(n+ by [email protected] » Thu Apr 15, 2021 10:29 am Vincen wrote: Wed Mar 31, 2021 8:13 am 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) $$\dfrac19$$

B) $$\dfrac13$$

C) $$\dfrac12$$

D) $$\dfrac23$$

E) $$\dfrac56$$

Answer: D

Source: GMAT Prep
Solution:

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

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