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Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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20 Mar 2012, 19:00

1

This post received KUDOS

to solve this problem: 1) substitute the constraint 2k+1 in the equation

A: (2k+1)(2k+2)(2k)/4 = (k+1)(k)(2k+1) DIVISIBLE by 4. B: (2k+1)(2k+2)/6 = (2k+1)(k+1)/3. But we don't get any clue about N itself. So, how can we say that n^3-n is divisible by 4. If anything from looking at both equations we know only that 2k+2 is divisible by 2. To be divisible by 4 we need to two sets of 2's which is lacking from the information given in B.

To answer your doubt: 1) n^2 + n is div by 6 then n^3-n is div by 4

Let's try: n = 2 it satisfies n^2 + n (6) is div by 6; BUT n^3-n(6) is NOT div by 4 (because 6 IS NOT divisible by 4). Let's try: n = 3 it satisfies n^2 + n(12) is div by 6; BUT n^3-n(24) IS div by 4 (because 24 IS divisible by 4).

So, since n= 2 doesnt work and n = 3 works. It's insufficient.

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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20 Mar 2012, 22:52

From Statement I, n = 2k+1, therefore n is odd. n^3-n = even.odd.even. Since there are atlest 2 powers of 2, it is divisble by 4.Sufficient

Statement II : n?^2+n = n(n+1).. values of n that satisfy this are 2,3,5,6.. however on substitution these value sin the question. the statemtnis tre for some and false for some eg,, when n = 2. n^3+n is divisible by 2 but not 4.. Hence insufficient

vix wrote:

Please help me with this...

IMHO whenever n^2 + n is div by 6 then n^3-n is div by 4. I checked this with 2,3,5. I dunno where I went wrong.

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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23 Mar 2012, 04:54

if n=1 though then the result is zero, is zero considered to be divisible by 4 ? so can we say zero is also a multiple of every integer ? if not , A is not sufficient by itself then.

if n=1 though then the result is zero, is zero considered to be divisible by 4 ? so can we say zero is also a multiple of every integer ? if not , A is not sufficient by itself then.

Zero is divisible by EVERY integer except zero itself, since 0/integer=integer (or, which is the same, 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]

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18 Apr 2012, 04:11

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?

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]

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15 Jun 2012, 15:37

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).

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]

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15 Aug 2013, 21:14

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) !
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Re: If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]

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08 Feb 2015, 01:23

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Re: If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]

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08 Feb 2015, 03:50

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.

Re: If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]

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22 Apr 2015, 15:39

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

St.1 n=2k+1. Plug in into equation above: (2k+1)(2k+2)2k=4k(2k+1)(k+1). Clearly divisible by 4. St.2 n^2 + n=n(n+1). If this equation is divisible by 6 then it must be divisible by 2 and 3. Hence we have to find if (n-1) is divisible by 2. If n=2 answer is no. If n=3 then answers is Yes. Hence insufficient. Answer A
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Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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18 Jun 2015, 05:30

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Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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05 Oct 2016, 22:05

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