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# DS-Divisible by 4.jpg

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Manager
Joined: 21 May 2008
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DS-Divisible by 4.jpg [#permalink]

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10 Oct 2008, 06:46
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Director
Joined: 27 Jun 2008
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WE 1: Investment Banking - 6yrs
Re: DS-Divisible by 4 [#permalink]

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10 Oct 2008, 07:35
(1) k = 2
n = 5

n^3-n = 125 - 5 = 120.....Yes

k=1
n=3

n^3 - n = 27-3 = 24...Yes

Suff

(2) n^2+n = n(n+1) / 6

Let n = 2

n^3 - n = 2^3-2 = 6...not divisible by 4

n = 5

n^3 - n = 120..divisible by 4

Insuff

A

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Senior Manager
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Re: DS-Divisible by 4 [#permalink]

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10 Oct 2008, 07:36
klb15 wrote:
Attachment:
DS-Divisible by 4.jpg

no easy way here

started from plugin in 1 and that eliminated 1)
then proceeded with 2 and with all the combinations it turned out that only numbers divisible w 6 but greater then 6 will satisfy, therefore C
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Senior Manager
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Posts: 372

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Re: DS-Divisible by 4 [#permalink]

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10 Oct 2008, 07:43
pawan203 wrote:
(1) k = 2
n = 5

n^3-n = 125 - 5 = 120.....Yes

k=1
n=3

n^3 - n = 27-3 = 24...Yes

Suff

(2) n^2+n = n(n+1) / 6

Let n = 2

n^3 - n = 2^3-2 = 6...not divisible by 4

n = 5

n^3 - n = 120..divisible by 4

Insuff

A

if n is 2 then satisfies 1) but not the question 8-2=6 not divisible
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VP
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Re: DS-Divisible by 4 [#permalink]

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10 Oct 2008, 07:50
I guess there is an easy way here.

$$n^3 -n$$ = $$n(n^2-1)$$ = n(n+1)(n-1)

The question is is (n-1)*n*(n+1) divisible by 4 ? ( notice n-1, n and n+1 are consecutive numbers)

(1) n=2k+1 . This essentially means n is odd. so n-1 and n+1 are evens.

Rule: When an integer can be divided by 2 twice, it is divisible by 4 as well.

Now as we have 2 evens, it is divisible by 2 twice or in other words it is divisible by 4.
So statement 1 is sufficient

(2) $$n^2+n$$ is divisible by 6 or n(n+1) is divisible by 6.
So either n or n+1 is a multiple of 2 and the other is multiple of 3. To determine whether a number is divisible by 4, we have to check if it is divisible by 2 twice. Now in statement 2 we do not know if n is even or odd. So cant say

Is the answer A ?
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Manager
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Re: DS-Divisible by 4 [#permalink]

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10 Oct 2008, 13:17
i am also getting answer A

i did it by picking odd and even numbers.

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SVP
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Re: DS-Divisible by 4 [#permalink]

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10 Oct 2008, 16:05
N(N+1)(N-1) ARE 3 CONSECS

IF N IS ODD THEN SURE IS DEVISBALE BY 4 IF N IS EVEN THEN IT COULD BE EITHER DEVISABLE BY 2 OR 4

FROM ONE

n IS ODD.......SUFF

FROM 2

N(N+1) IS DEVISBALE BY 6

WE STILL DONT KNOW IF N IS EVEN OR ODD.....INSUFF

a

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Manager
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Re: DS-Divisible by 4 [#permalink]

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11 Oct 2008, 11:29
Quest is n3-n is divisible by 4
OR
n(n-1)(n+1) is divisible by 4.

Now state 1
------------

n = 2k + 1 => (n-1) = 2k means n-1 always divisible by 2

2nd state
---------
n^2 + n is divisible by 6.
=> n(n+1) is divisible by 6

so n(n-1)(n+1) is divisible by 12 Hence divisible bhy 4

Ans is C)

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Re: DS-Divisible by 4   [#permalink] 11 Oct 2008, 11:29
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# DS-Divisible by 4.jpg

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