N is an integer greater than 6, which of the following must

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N is an integer greater than 6, which of the following must [#permalink]

11 Jul 2007, 23:15

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N is an integer greater than 6, which of the following must be divisible by 3?

n (n+2) (n+3)
n (n+5) (n-3)
n (n+2) (n+5)
n (n+1) (n-4)
n (n+1) (n-2)

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12 Jul 2007, 00:34

D.

1st method (easiest):
n=8 then A,B.C. OUT
n=7-E out

2-method:

D because
D=n(n+1)(n-1-3)=n(n+1)[(n-1)-3]
= (n-1)n(n+1)-3n(n+1) is clearly divisible by 3.

Because n-1, n,n+1 is 3 consequitive #s. Each time at least one of them is always divisible by 3. And 3 is a factor of 3n(n+1)

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12 Jul 2007, 07:30

UMB wrote:

=n(n+1)(n-1-3)=n(n+1)[(n-1)-3]

excellent.

OA is D.

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12 Jul 2007, 10:44

UMB wrote:

Excellent confirmation. I used 8 and 7 to eliminate and get to D. But dd not try the confirmation

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12 Jul 2007, 11:24

This is a great example of a question that would appear in the first 5 questions on the math section. The gmat likes to put questions there that aren't necessarily hard, but are tricky. It's easy to lose track of what you are doing on a question like this and put down a wrong answer.

Take your time with these on test day. They count infinitely more than a probability or modulus question that appears toward the end of the section.

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Re: PS: Divisible by 3 [#permalink]

12 Jul 2007, 13:36

Himalayan wrote:

N is an integer greater than 6, which of the following must be divisible by 3?

n (n+2) (n+3)
n (n+5) (n-3)
n (n+2) (n+5)
n (n+1) (n-4)
n (n+1) (n-2)

Also, note that n-1 will have the same remainder when divided by 3 as n - 4, so n(n+1)(n-4) will have the same remainder as n(n+1)(n-1). Since n -1 ,n and n+1 are three consecutive integers, the remainder will be 0, i.e. the product is a mulltiple of 3

Re: PS: Divisible by 3 [#permalink] 12 Jul 2007, 13:36

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N is an integer greater than 6, which of the following must

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N is an integer greater than 6, which of the following must

N is an integer greater than 6, which of the following must

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