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# NUMBER PROPERTIES

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Director
Joined: 23 May 2008
Posts: 801

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13 Jun 2009, 15:51
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67% (00:43) correct 33% (00:56) wrong based on 5 sessions

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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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)

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SVP
Joined: 29 Aug 2007
Posts: 2472

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13 Jun 2009, 19:58
bigtreezl wrote:
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)

(A) n (n+1) (n-4) = n (n+1) (n-4+6) = n (n+1) (n+2)

This is equivalant to a multiplication of any 3 contineous integers > 6. In this case, the product (of any 3 contineous integers > 6)must be a multiple of 3.

(B) n (n+2) (n-1) = n (n+2) (n-1+3) = n (n+2) (n+2)

This is not equivalant to a multiplication of any 3 contineous integers > 6. In this case, the product may/may not be a multiple of 3.

(C) n (n+3) (n-5) = n (n+3) (n-5+6) = n (n+3) (n+1)

This is not equivalant to a multiplication of any 3 contineous integers > 6. In this case, the product may/may not be a multiple of 3.

(D) n (n+4) (n-2) = n (n+4) (n-2+3)= n (n+4) (n+1)

This is not equivalant to a multiplication of any 3 contineous integers > 6. In this case, the product may/may not be a multiple of 3.

(E) n (n+5) (n-6) = n (n+5) (n-6+9) = n (n+5) (n+3).

This is not equivalant to a multiplication of any 3 contineous integers > 6. In this case, the product may/may not be a multiple of 3.

Then its A.
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Manager
Joined: 28 Jan 2004
Posts: 202

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Location: India

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13 Jun 2009, 20:02
best way to solve this is by substituting numbers else it will be very combersome.

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Manager
Joined: 28 Jan 2004
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13 Jun 2009, 20:06
Gmattiger can ypu pls. explain your below stmt.
A) n (n+1) (n-4) = n (n+1) (n-4+6) = n (n+1) (n+2)

This is equivalant to a multiplication of any 3 contineous integers > 6. In this case, the product (of any 3 contineous integers > 6)must be a multiple of 3.

How can you say that n-4 is equivalent to n+2 ?

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SVP
Joined: 29 Aug 2007
Posts: 2472

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13 Jun 2009, 20:16
1
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mdfrahim wrote:
Gmattiger can ypu pls. explain your below stmt.
A) n (n+1) (n-4) = n (n+1) (n-4+6) = n (n+1) (n+2)

This is equivalant to a multiplication of any 3 contineous integers > 6. In this case, the product (of any 3 contineous integers > 6)must be a multiple of 3.

How can you say that n-4 is equivalent to n+2 ?

Thats the point here.
Suppose n is any integer > 6. Lets say n = 13, then (n - 4) = 9 and (n +2) = 15. If (n+2) is a multiple of 3, then (n-4) is also a multiple of 3. In terms of multiples of 3, (n-4) is equivalant to (n+2).
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Manager
Joined: 28 Jan 2004
Posts: 202

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14 Jun 2009, 23:04
Thx. Gmat tiger. I got it now.

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Re: NUMBER PROPERTIES   [#permalink] 14 Jun 2009, 23:04
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