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

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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

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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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