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# An integer n between 1 and 99, inclusive, is to be chosen at

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Kudos [?]: 2 [0], given: 18

An integer n between 1 and 99, inclusive, is to be chosen at [#permalink]  04 Oct 2013, 02:12
00:00

Difficulty:

35% (medium)

Question Stats:

58% (02:38) correct 41% (01:12) wrong based on 65 sessions
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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Nov 2013, 08:54, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
Math Expert
Joined: 02 Sep 2009
Posts: 16878
Followers: 2776

Kudos [?]: 17638 [6] , given: 2233

Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink]  04 Oct 2013, 03:34
6
KUDOS
Expert's post
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

n(n+1) to be divisible by 3 either n or n+1 must be a multiples of 3.

In each following group of numbers: {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, ..., {97, 98, 99} there are EXACTLY 2 numbers out of 3 satisfying the above condition. For example in {1, 2, 3} n can be: 2, or 3. Thus, the overall probability is 2/3.

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Kudos [?]: 106 [2] , given: 72

Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink]  26 Dec 2013, 22:54
2
KUDOS
the required number = (if n is even )+ (if n is divisible by 3) - (if n is even and divisible by 3)=
=(49+33-16)/99=66/99=2/3
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Re: An integer n between 1 and 99, inclusive, is to be chosen at   [#permalink] 26 Dec 2013, 22:54
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