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# If n is an integer greater than 6, which of the following

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VP
Joined: 22 Nov 2007
Posts: 1080
If n is an integer greater than 6, which of the following [#permalink]

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07 Mar 2008, 12:45
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If 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)
Director
Joined: 10 Sep 2007
Posts: 938

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07 Mar 2008, 13:36
If n > 6 then there are 6 possibilities for n,

n = 6k + 1
n = 6k + 2
n = 6k + 3
n = 6k + 4
n = 6k + 5
n = 6k + 6

Here K is any integer greater than or equal to 1. For simplicity we take K =1, so n can be 7, 8, 9, 10, 11, 12

Now we need to put these possibilities in the answers given.

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)

For n = 7, values are
7*9*10
7*12*4
7*9*12
7*8*3
7*8*5

Note than E is not divisible by 3, so we will continue with other values only for first 4 options.

For n = 8, values are
8*10*11
8*13*5
8*10*13
8*9*4
8*9*6
Again note that A, & B are not divisible by 3 so they are out as well.

Now for 9, value are
9*10*5
9*10*7

As no conclusive answer so we proceed with 10, and values are
10*11*6
10*11*8

So we see that D is out only answer remaining is D.

Just to verify that it is the answer we put remaining 2 values (11, & 12) in it to verify 100%.
11*12*7

12*13*8

Both are divisible by 3 so D is the answer.
VP
Joined: 22 Nov 2007
Posts: 1080

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08 Mar 2008, 10:21
abhijit_sen wrote:
If n > 6 then there are 6 possibilities for n,

n = 6k + 1
n = 6k + 2
n = 6k + 3
n = 6k + 4
n = 6k + 5
n = 6k + 6

Here K is any integer greater than or equal to 1. For simplicity we take K =1, so n can be 7, 8, 9, 10, 11, 12

Now we need to put these possibilities in the answers given.

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)

For n = 7, values are
7*9*10
7*12*4
7*9*12
7*8*3
7*8*5

Note than E is not divisible by 3, so we will continue with other values only for first 4 options.

For n = 8, values are
8*10*11
8*13*5
8*10*13
8*9*4
8*9*6
Again note that A, & B are not divisible by 3 so they are out as well.

Now for 9, value are
9*10*5
9*10*7

As no conclusive answer so we proceed with 10, and values are
10*11*6
10*11*8

So we see that D is out only answer remaining is D.

Just to verify that it is the answer we put remaining 2 values (11, & 12) in it to verify 100%.
11*12*7

12*13*8

Both are divisible by 3 so D is the answer.

your answer is right. anyway, even though using examples can be useful I get often confused and lose time. I reasoned in this way:

we know that the pruduct of any 3 consecutive numbers is a multiple of 3. thus,

(n-1)n(n+1) is a multiple of 3. unfortunately gmat would make it so easy. anyway we can think that, even though we subsituted (n-1) with (n-4) the results wouldn't change the divisibility of (n-1)n(n+1) by 3. thus OA is D
Re: integer gmatprep   [#permalink] 08 Mar 2008, 10:21
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