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# NUMBER THEORY

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Manager
Joined: 04 Sep 2006
Posts: 113

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02 Feb 2009, 11:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Q : N IS AN INTEGER FROM 1 TO 99 INCLUSIVVE WHAT IS THE PROB THAT N(N+1) IS DIVISIBLE BY 3 ?

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SVP
Joined: 29 Aug 2007
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02 Feb 2009, 12:31
vcbabu wrote:
Q : N IS AN INTEGER FROM 1 TO 99 INCLUSIVVE WHAT IS THE PROB THAT N(N+1) IS DIVISIBLE BY 3 ?

if n = 1, n(n+1) = 1x2 ...... not divisible.
if n = 2, n(n+1) = 2x3 ...... divisible.
if n = 3, n(n+1) = 3x4 ...... divisible.
if n = 4, n(n+1) = 4x5 ...... not divisible.
if n = 5, n(n+1) = 5x6 ...... divisible.
if n = 6, n(n+1) = 6x7 ..... divisible.
if n = 7, n(n+1) = 7x8 ...... not divisible.

so on.

the pattren is 2 out of 3 is divisible by 3. so the prob = 2/3 or 66.67%.
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Kudos [?]: 843 [0], given: 19

Senior Manager
Joined: 30 Nov 2008
Posts: 485

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Schools: Fuqua

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02 Feb 2009, 13:10
1
KUDOS
Agree that the prpbability is 2/3. If we have to go by probability theory, then here is my approach.

Total no of numbers between 1 - 99 = 99.
Event is picking a number whose n(n+1) is divisible by 3 which means either n should be divisible by 3 or n+1 is divisible by 3.
Total Number of numbers from 1 to 99 divisisible by 3 is 33.
Total number of number from 1 to 99 where n+1 is divisible by 3 is also 33.

Probability = No of possible events / total number of events = 66 / 99 = 2/3.

Kudos [?]: 361 [1], given: 15

Re: NUMBER THEORY   [#permalink] 02 Feb 2009, 13:10
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# NUMBER THEORY

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