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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 _____________________________________________________________________________

gmatclubot

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