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Re: If n is a positive integer, is n^3 n divisible by 4 ? (1) n [#permalink]
18 Apr 2012, 03:11

In the statement one when we express n=2k+1, why cant we take k=0, in that case n=1 and the product will be zero...Can we say then zero is divisible by 4?

Re: If n is a positive integer, is n^3 n divisible by 4 ? (1) n [#permalink]
18 Apr 2012, 03:14

Expert's post

1

This post was BOOKMARKED

pavanpuneet wrote:

In the statement one when we express n=2k+1, why cant we take k=0, in that case n=1 and the product will be zero...Can we say then zero is divisible by 4?

Zero is a divisible by every integer, except zero itself. Or which is the same: zero is a multiple of every integer, except zero itself. Integer \(a\) is a multiple of integer \(b\) means that \(a\) is "evenly divisible" by \(b\), i.e., divisible by \(b\) without a remainder. Now, since zero/integer=integer then zero is a multiple of every integer (except zero itself).

Re: If n is a positive integer, is n^3 n divisible by 4 ? (1) n [#permalink]
15 Jun 2012, 14:37

Bunuel wrote:

pavanpuneet wrote:

In the statement one when we express n=2k+1, why cant we take k=0, in that case n=1 and the product will be zero...Can we say then zero is divisible by 4?

Zero is a divisible by every integer, except zero itself. Or which is the same: zero is a multiple of every integer, except zero itself. Integer \(a\) is a multiple of integer \(b\) means that \(a\) is "evenly divisible" by \(b\), i.e., divisible by \(b\) without a remainder. Now, since zero/integer=integer then zero is a multiple of every integer (except zero itself).

Re: If n is a positive integer, is n^3 n divisible by 4 ? (1) n [#permalink]
15 Jun 2012, 18:02

1

This post received KUDOS

Expert's post

solarzj wrote:

Bunuel wrote:

pavanpuneet wrote:

In the statement one when we express n=2k+1, why cant we take k=0, in that case n=1 and the product will be zero...Can we say then zero is divisible by 4?

Zero is a divisible by every integer, except zero itself. Or which is the same: zero is a multiple of every integer, except zero itself. Integer \(a\) is a multiple of integer \(b\) means that \(a\) is "evenly divisible" by \(b\), i.e., divisible by \(b\) without a remainder. Now, since zero/integer=integer then zero is a multiple of every integer (except zero itself).

Re: If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]
15 Aug 2013, 20:14

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

REM(X/4)=?

(1).

n is odd => For every odd number there exists two even numbers in the expression (n-1) and (n+1) Hence divisible by 4 . SUFFICIENT

(2).

n*(n+1) is DIV by 6

n=2 => X=1*2*3 = 6 REM(X/4)=> REM(6/4) = 2

n=3 => X=2*3*4 = 24 REM(X/4) => REM(24/4) = 0

Hence two remainders on two cases => INSUFFICIENT

Hence (A) ! _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]
08 Feb 2015, 00:23

Hello from the GMAT Club BumpBot!

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Re: If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]
08 Feb 2015, 02:50

Baten80 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.

(n^3 - n) mod 4 = 0 ? n(n^2 - 1) mod 4 = 0 ? (n-1)(n)(n+1) mod 4 = 0 ?

1)n = 2k+1 => 4k(2k+1)(k+1) this is div by 4. hence, sufficient.

2) n^2 + n mod 6 = 0 n(n+1) mod 6 = 0 so either n or n+1 must be divisible by 2 if n is div by 2 then original expression is not div by 4 if n+1 is div by 2 then original expression is div by 4 hence, insufficient.

A. _________________

Illegitimi non carborundum.

gmatclubot

Re: If n is a positive integer, is n^3 n divisible by 4 ?
[#permalink]
08 Feb 2015, 02:50

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