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If n is a positive integer, is n^3 - n divisible by 4? 1) n

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Director
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If n is a positive integer, is n^3 - n divisible by 4? 1) n [#permalink] New post 27 Jul 2007, 20:55
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If n is a positive integer, is n^3 - n divisible by 4?

1) n = 2k + 1, where k is an integer.

2) n^2 + n is divisible by 6.


Guys, in (1), can k be equal to 0?
Director
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Re: DS - OG 147 [#permalink] New post 27 Jul 2007, 21:52
asaf wrote:
If n is a positive integer, is n^3 - n divisible by 4?

1) n = 2k + 1, where k is an integer.

2) n^2 + n is divisible by 6.


Guys, in (1), can k be equal to 0?


Well the OA is A.

For 1): If k = 0; then n = 1 ==> n^3 - n = 0 which is not divisible by 4, right?

if n = 3 (k=1) then (n-1) n (n+1) = 2 * 3 * 4 is divisble by 4

Not sufficient, right?
why OA is A?
Senior Manager
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Re: DS - OG 147 [#permalink] New post 27 Jul 2007, 22:51
asaf wrote:
asaf wrote:
If n is a positive integer, is n^3 - n divisible by 4?

1) n = 2k + 1, where k is an integer.

2) n^2 + n is divisible by 6.


Guys, in (1), can k be equal to 0?


Well the OA is A.

For 1): If k = 0; then n = 1 ==> n^3 - n = 0 which is not divisible by 4, right?

if n = 3 (k=1) then (n-1) n (n+1) = 2 * 3 * 4 is divisible by 4

Not sufficient, right?
why OA is A?


I think 0 is divisible by all numbers. So A should be sufficient.
VP
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 [#permalink] New post 28 Jul 2007, 01:05
statement 1

n = 2k + 1, where k is an integer.

plug in for n

(2k + 1)^3 - (2k + 1)

(2k + 1)*((2k + 1)^2 - 1)

(2k + 1)*(4k^2 + 4k + 1 - 1)

(2k + 1)*(4k^2 + 4k)

8k^3 + 4k^2 + 4K + 8k^2

4(2k^3 + k^2 + k + 2k^2)

so we can say that n^3 - n divisible by 4

sufficient

0 is divisible by all numbers (see OG 11 page 108) n*0 = 0 so 4*0 = 0 and 0/4 = 0

statement 2

n^2 + n is divisible by 6.

clearly insufficient

the answer is (A)

:-D
Manager
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 [#permalink] New post 28 Jul 2007, 03:37
A is correct.
1. n=2k+1 i.e. n will always be odd and that will make n^3-n always even. And since it is sqared so it can be divided by 4
2. It is divided by 6 i.e. it can be divided by 2 and 3 so here n be or odd or even. Not sufficient
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 [#permalink] New post 28 Jul 2007, 05:11
KillerSquirrel wrote:
statement 1

n = 2k + 1, where k is an integer.

plug in for n

(2k + 1)^3 - (2k + 1)

(2k + 1)*((2k + 1)^2 - 1)

(2k + 1)*(4k^2 + 4k + 1 - 1)

(2k + 1)*(4k^2 + 4k)

8k^3 + 4k^2 + 4K + 8k^2

4(2k^3 + k^2 + k + 2k^2)

so we can say that n^3 - n divisible by 4

sufficient

0 is divisible by all numbers (see OG 11 page 108) n*0 = 0 so 4*0 = 0 and 0/4 = 0

statement 2

n^2 + n is divisible by 6.

clearly insufficient

the answer is (A)

:-D


Same approach except that, for the statment 1, I factorize to avoid huger calculus :)

n^3 - n
= n*(n^2 - 1)
= n*(n+1)*(n-1)
= (2*k+1) * ((2*k+1) +1) * ((2*k+1) -1)
= (2*k+1) * 2*(k+1) * 2*k
= 4 * (2k+1) * (k+1) * k

:)
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 [#permalink] New post 28 Jul 2007, 08:21
KillerSquirrel wrote:
statement 1

n = 2k + 1, where k is an integer.

plug in for n

(2k + 1)^3 - (2k + 1)
(2k + 1)*((2k + 1)^2 - 1)
(2k + 1)*(4k^2 + 4k + 1 - 1)
(2k + 1)*(4k^2 + 4k)
8k^3 + 4k^2 + 4K + 8k^2
4(2k^3 + k^2 + k + 2k^2)

so we can say that n^3 - n divisible by 4

sufficient

0 is divisible by all numbers (see OG 11 page 108) n*0 = 0 so 4*0 = 0 and 0/4 = 0

statement 2

n^2 + n is divisible by 6.

clearly insufficient

the answer is (A)

:-D



n = 2k + 1, where k is an integer. plug in for n

(2k + 1)^3 - (2k + 1)
(2k + 1)*((2k + 1)^2 - 1)
(2k + 1)*(4k^2 + 4k + 1 - 1)
(2k + 1)*(4k^2 + 4k)
(2k + 1)* 4 (k^2 + k)
gotta. so I stop here.
VP
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 [#permalink] New post 28 Jul 2007, 09:58
Himalayan wrote:

n = 2k + 1, where k is an integer. plug in for n

(2k + 1)^3 - (2k + 1)
(2k + 1)*((2k + 1)^2 - 1)
(2k + 1)*(4k^2 + 4k + 1 - 1)
(2k + 1)*(4k^2 + 4k)
(2k + 1)* 4 (k^2 + k)
gotta. so I stop here.


Perfect - thanks

:-D
  [#permalink] 28 Jul 2007, 09:58
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