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 Post subject: Dividing by 3
PostPosted: Wed Apr 29, 2009 3:34 pm 

Joined: Sun Apr 19, 2009 6:10 pm
Posts: 3
Hey guys, I came across a problem that I could not understand. Is there anyone that can help me out. This thing is driving me crazy.

Given that n is an integer, is n-1 divisible by 3 ?

(1) n^2 + n is not divisible by 3
(2) 3n + 5 = or > k + 8 where k is a multiple of 3

 Post subject: Re: Dividing by 3
PostPosted: Fri May 01, 2009 10:12 pm 
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Joined: Mon Apr 06, 2009 5:44 pm
Posts: 81
Hey Ewok,

Lets begin with some notation. Let # mean not equal to and let > mean greater or equal to.

(1) can be rewritten in the following way n(n+1) # 3. It should be clear that n and n+1 are consecutive numbers and that any three consecutive numbers must contain at least one that has a factor of 3 in it. So, according to our statement both n-1 (and n+2) must be divisible by 3. It works for negative numbers and zero as well. Remember, zero is divisible by 3 since 0/3 = 0. And, neither n or n+1 can be zero since that would make their product divisible by 3. A is sufficient.

Lets look at (2). We can rewrite 3n+5 > k +8. Since k is a multiple of 3, lets substitute 3x for k where x is any non-negative integer, and by subtracting 5 from both sides: 3n > 3x + 3, which implies that n > x + 1. This tells us nothing about the divisibility of n-1. It merely tells us that n-1 is non-negative.

The correct answer is A.

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