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# OG 12th Edition - Problem Solving, Question 82 - Help

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OG 12th Edition - Problem Solving, Question 82 - Help [#permalink]

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04 Nov 2009, 12:58
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If n is an integer greater than 6, which of the following must be divisible by 3?

A) N(N+1)(N-4)
B) N(N+2)(N-1)
C) N(N+3)(N-5)
D) N(N+4)(N-2)
E) N(N+5)(N-6)

Aside from plugging numbers what is a better algebraic strategy? Or is plugging numbers the quickest? I alwasy get flustered and unsure of which number to start with so maybe there is a more direct approach.

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Re: OG 12th Edition - Problem Solving, Question 82 - Help [#permalink]

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04 Nov 2009, 13:42
The strategy is more intuitive than anything. Since 3 is a prime number, you know that if any of those individual terms are divisible by 3 (i.e. N, N+1, etc.), than the whole product will be divisible by 3.

The thought process I use is as follows: essentially (without using algebraic expansion) you need to look for 3 consecutive numbers. The reasoning is that if you have N * (N+1) * (N+2), one of the terms must be divisible by 3. None of the cases at first glance fit this criteria.

The next step is the key: you need to determine which answer has 3 terms that would have 3 different remainders when divided by 3. For example, take E: N(N+5)(N-6). The term (N-6) would have the same remainder as the term N, since they have a difference of 6 (also divisible by 3).

Just by looking at the answers mentally, A: N(N+1)(N-4) is the only answer that satisfies this.
Re: OG 12th Edition - Problem Solving, Question 82 - Help   [#permalink] 04 Nov 2009, 13:42
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# OG 12th Edition - Problem Solving, Question 82 - Help

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