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

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If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink] New post 20 Jan 2007, 21:26
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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.





OG has a convoluted explanation that gives (A) as the answer.

However I solve it using plug-in method and get (E). Could someone please tell me where I made a mistake.


Statement 1

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

As per statement 1, If K is an integer, then K can be -2 , -1, 0 , 1, 2, 3

When K = 0 ( based on problem statement, K is an integer )
n = 2 ( 0) + 1 = 1

When K = - 1 ( based on problem statement, K is an integer )
n = 2 ( - 1) + 1 = - 1

When k = 1
n = 2 ( 1) + 1 = 3

Taking values of K where n is a positive integer

If K = 0 , then n = 1 , and n^3 – n = 0

Therefore for statement 1 when K is the 0, the statement is insufficient for n^3 – n to be divisible by 4.

Statement 1 is sufficient only when k is a positive integer greater than 0 , a condition not explicitly given in statement 1.


Statement 2

Insufficient.


Both Statement 1 and 2 are insufficient when taken together

Therefore (E)
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Re: OG DS - Don't understand the answer [#permalink] New post 21 Jan 2007, 01:57
focused07 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.





OG has a convoluted explanation that gives (A) as the answer.

However I solve it using plug-in method and get (E). Could someone please tell me where I made a mistake.


Statement 1

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

As per statement 1, If K is an integer, then K can be -2 , -1, 0 , 1, 2, 3

When K = 0 ( based on problem statement, K is an integer )
n = 2 ( 0) + 1 = 1

When K = - 1 ( based on problem statement, K is an integer )
n = 2 ( - 1) + 1 = - 1

When k = 1
n = 2 ( 1) + 1 = 3

Taking values of K where n is a positive integer

If K = 0 , then n = 1 , and n^3 – n = 0

Therefore for statement 1 when K is the 0, the statement is insufficient for n^3 – n to be divisible by 4.

Statement 1 is sufficient only when k is a positive integer greater than 0 , a condition not explicitly given in statement 1.


Statement 2

Insufficient.


Both Statement 1 and 2 are insufficient when taken together

Therefore (E)


In bold, I show the problems in your reasoning :)

In the first bold, k cannot be negative because n is a positive integer, definied such as this. :)

In the second bold, 0/4 = 0. 0 is disivable by any integer. Why? For any integer a, we have 0/a = 0*k + 0. The remainder is always equal to 0, the definition of divisibility :)

Last edited by Fig on 21 Jan 2007, 03:40, edited 3 times in total.
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 [#permalink] New post 21 Jan 2007, 03:31
To do DS divisibility and remainder problems, it really helps to understand how to write and interpret things written in mod form, where:

n = dk + r, where n is an integer, d is an integer divisor, k is any integer, and r is an integer remainder

Examples:

x is divisible by 3 can be written: x = 3k + 0, for any integer k
x, when divided by 3 has a remainder of 1: x = 3k + 1, for any integer k

In this problem, statement one states:

n = 2k + 1, for any integer k. What this means is that when n is divided by 2, there is a remainder of 1. This in turn is just a convoluted way of saying n is an odd integer.

If you plug any odd integars into the formual n^3 - n, you'll find that they are divisible by four.
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 [#permalink] New post 21 Jan 2007, 08:26
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.


n^3 - n = n(n^2 - 1) = n(n-1)(n+1) ie the product of 3 consecutive intigers

to be devisble by 4 we have to ve one of 2 scenarios

1) n is odd thus n-1 , n+1 are even and thus their product is devisble by 4

or

2) n is even and devisble by 4 ( not just even because 2 is even but not devisble by 4)

from one

n is odd in the form of 2K + 1 ( we know even + odd = odd) thus

scenario one is valid ..........suff

from two

n(n+1) is devible by 6 , means if n is odd and a multiple of 3 thus n+1 is even and vica versa

and we can never deduce if the even component of n ( n+1) is devisble by 4 as it could be only devisible by 2)

i hope this helps

My answer is A
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Re: OG DS - Don't understand the answer [#permalink] New post 21 Jan 2007, 09:53
[quote="focused07"]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.

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

given n^2 + n is div by 6 then either n or n^2+1 is div by 2 we can say that n^3 - n is div by 4 only for the case where n is odd

from (1)
if n = 2k + 1 we have (2k+1)*2k*(2k+2) = 4*(2k+1)*k*(k+1) so 1 is suff


so A ?
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 [#permalink] New post 22 Jan 2007, 02:44
Is n^3 - n divisible by 4.
=>Is n(n^2 - 1) divisible by 4.
(1) says: n = 2k + 1

Put (1) into the question:
=> (2k + 1)* ( (2k + 1)^2 - 1)
= (2k + 1)*( 4k^2 + 1 + 4k - 1)
= (2k + 1)*(4k^2 + 4k)
= 8k^3 + 4k^2 + 8k + 4k

All terms are divisible by 4.
Thus n^3 - n is divisible by 4.

A is sufficient.
However, ProfessorMMA's explanation is much better. :-D
  [#permalink] 22 Jan 2007, 02:44
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