If n is an integer greater than 6, which of the following : PS Archive
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If n is an integer greater than 6, which of the following

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

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27 Mar 2007, 06:11
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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)

I'm starting to like these...
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27 Mar 2007, 07:34
got A. did not go past choice A. is there an easier way than substituting numbers.
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27 Mar 2007, 07:45
(A) by picking numbers. I think it must be the easiest way for this problem
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27 Mar 2007, 09:40
The correct expression will be the one which is true for every integer :

Here for e.g the three nos are n ,n+1,n-4

Now if the sum of these numbers is a multiple of 3 ,then the original expression will be always valid.

For
A Sum =3n-3 .....is a multiple
B Sum = 3n+1 ..Not
C Sum = 3n-2..Not
D Sum= 3n +2..Not
E. Sum= 3n-1 ..Not a multiple.

Check: 3n+3 will be valid expression i.e (n)(n-1)(n+4)

Personally, I wud have picked numbers (which I usually hate!)...
Is this a GMATPrep Q ..I doubt

Last edited by suithink on 27 Mar 2007, 17:41, edited 1 time in total.
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27 Mar 2007, 13:04
(A) for me as well

To me, we have to find an expression that is similar to:
n * (n+1 +3*k) * (n+2 +3*j) where k and j are integers.

An example is:
n * (n+1) * (n+2) that is always divisable by 3 no matter the value of n

Thus,

A) Bingo !
n(n+1)(n-4) = n(n+1)(n+2 -3*2) (k=0 and l=-2)

B) No
n(n+2)(n-1)
= n(n+1 -2/3*3)(n+2) (k=-2/3 : not an integer and l=0)

C) No
n(n+3)(n-5)
= n(n+1 -3*2)(n+2 +1/3*3) (k=-2 and l=1/3 : not an integer)

D) No
n(n+4)(n-2)
= n(n+1 +3*1)(n+2 -4/3*3) (k=-2 and l=-4/3 : not an integer)

E) No
n(n+5)(n-6)
= n(n+1 -7/3*3)(n+2 +3*1) (k=-7/3 : not an integer and l=3)
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27 Mar 2007, 17:12
I have tried plugging in numbers.

Is this the rule - if the sum of the three integers is divisible by 3, the product will also be divisible by 3. Though it looks like true for all combinations I have tried.
Can somebody give the authentic reference for that.
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27 Mar 2007, 17:13
I have tried plugging in numbers.

Is this the rule - if the sum of the three integers is divisible by 3, the product will also be divisible by 3. Though it looks like true for all combinations I have tried.
Can somebody give the authentic reference for that.
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28 Mar 2007, 05:08
The correct expression will be the one which is true for every integer :

Here for e.g the three nos are n ,n+1,n-4

Now if the sum of these numbers is a multiple of 3 ,then the original expression will be always valid.

Let the given numbers be arranged in a matrix ...

(n-6) (n-5) (n-4)
(n-3) ( n-2) (n-1)
n (n+1) (n+2)
n+3 n+3 n+4

In each row 'one ' number will be a multiple of 3...as in three consecutive numbers one WILL BE a multiple of 3.Also if for e.g (n-4) is a multiple of 3 then all terms in 3rd Column will also be multiple of 3.
==>Those Answer choices which has numbers from each column will be 'definitely' a multiple.
Or
T==>he sum of elements in each row will be a multiple of 3...!

For
A Sum =3n-3 .....is a multiple
B Sum = 3n+1 ..Not
C Sum = 3n-2..Not
D Sum= 3n +2..Not
E. Sum= 3n-1 ..Not a multiple.
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28 Mar 2007, 06:43
Good work!

The OA is A.
28 Mar 2007, 06:43
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