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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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Wed Mar 31, 2021 8:13 am

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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 Scott@TargetTestPrep » 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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