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

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

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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.
[Reveal] Spoiler: OA

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Baten80 wrote:
170. If n is a positive integer, is n3 – n divisible by 4 ?
(1) n = 2k + 1, where k is an integer.
(2) n2 + n is divisible by 6.


If n is a positive integer, is n^3 – n divisible by 4 ?

n^3-n=n(n^2-1)=(n-1)n(n+1), so we are asked whether the product of 3 consecutive integers is divisible bu 4.

(1) n = 2k + 1, where k is an integer --> n=odd --> as n is odd then both n-1 and n+1 are even hence (n-1)n(n+1) is divisible by 4. Sufficient.

(2) n^2 + n is divisible by 6 --> if n=2 then n^3-n=6 and the answer is NO but if n=3 then n^3-n=24 and the answer is YES. Not sufficient.

Answer: A.

P. S. Baten80 please format the questions properly.
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If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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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

Last edited by Bunuel on 20 Nov 2014, 07:53, edited 2 times in total.
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Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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Impenetrable 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

I figured out that 1) is enough...but how do I figure out that 2) is sufficient/ not sufficient?
Any tipps for divisibility questions?


Cheers!


If n is a positive integer, is n^3 – n divisible by 4 ?

n^3-n=n(n^2-1)=(n-1)n(n+1), so we are asked whether the product of 3 consecutive integers is divisible bu 4.

(1) n = 2k + 1, where k is an integer --> n=odd --> as n is odd then both n-1 and n+1 are even hence (n-1)n(n+1) is divisible by 4. Sufficient.

(2) n^2 + n is divisible by 6 --> if n=2 then n^3-n=6 and the answer is NO but if n=3 then n^3-n=24 and the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.
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Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]

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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.

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

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New post 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.

many thanks...

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

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New post 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.

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

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New post 23 Mar 2012, 05:01
goktug88 wrote:
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).

Check Number Theory chapter of Math Book for more hints/tips/rules on this subject: math-number-theory-88376.html

Hope it helps.
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Re: If n is a positive integer, is n^3 n divisible by 4 ? (1) n [#permalink]

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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?

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

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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).

Check Number Theory chapter of Math Book for more: math-number-theory-88376.html

Hope it helps.
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If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]

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New post 10 Jun 2012, 14:18
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


Vote for A

(A) n = 2K + 1 therefore n = {1,3,5,7,9,11 ......}

when n=3 then 3(9-1) is divisible by 4 sufficient
all values of n = {1,3,5,7,9,11 ......} will be divisiable by 4

Information sufficent

(B) n^2 + n is divisible by 6

n(n+1) = 6q (q = any multiple of 6)
n & (n+1) are +ve consecutive integers and therefore co-prime numbers

therefore when
n=6 (n+1) = 7 divisible by 6 but not by 4
n=12 (n+1) = 13 divisible by 6 and 4

Information not sufficent

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

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New post 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).

Check Number Theory chapter of Math Book for more: math-number-theory-88376.html

Hope it helps.


Sorry Bunuel, just to make this clear

if k = 0 then n = 1 (2 * 0 + 1)

this means that (n-1)(n)(n+1) is 0 * 1 * 2

which is 0 (zero) and therefore divisible by 4 because zero is divisible by 4 ... right?
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Re: If n is a positive integer, is n^3 n divisible by 4 ? (1) n [#permalink]

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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).

Check Number Theory chapter of Math Book for more: math-number-theory-88376.html

Hope it helps.


Sorry Bunuel, just to make this clear

if k = 0 then n = 1 (2 * 0 + 1)

this means that (n-1)(n)(n+1) is 0 * 1 * 2

which is 0 (zero) and therefore divisible by 4 because zero is divisible by 4 ... right?


Exactly so: if n=1 then (n-1)(n)(n+1)=0*1*2=0 and 0 is divisible by every positive integer including 4.
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Re: If n is a positive integer, is n^3 n divisible by 4 ? [#permalink]

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New post 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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New post 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.

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

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

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New post 03 Mar 2016, 19:45
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.


for n^3 -n to be divisible by 4, we need to know for sure whether n is even or odd.

1 - n is odd - so sufficient.

2. 2 options:
n=3; n+1=4 => yes
n=2; n+1=3 => no

2 outcomes - insufficient.

A.

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

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New post 15 Jan 2017, 01:01
Excellent Official Question.
Here is what i did in this one =>

Firstly n^3-n= (n-1)n(n+1)
It is a product of 3 consecutive integers.
Regardless of what the value of n is => It will always be a multiple of 6.
Reason -> Product of n consecutive integers is divisible by n!

Now lets see two cases ->

Case 1-> n is odd
If n is odd => n-1 will be even and n+1 will be even too.
Further more one out of them will be multiple of 4.
Thus (n-1)n(n+1) will be multiple of 8.
As it is already a multiple of 6 => we can conclude that (n-1)n(n+1) will be a multiple of 24.

Case 2-> n is even
In that scenario it is not possible for us to conclude anything except (n-1)n(n+1) will be multiple of 6.



Hence we just need to see if n is odd or not.

Statement 1->
n=2k+1 for some integer k.
Hence n=even+odd=odd
Hence n must be a multiple of 24.
Also it will be a multiple of 4.

Hence sufficient.

Statement 2->
Lest use some test cases=>
n=2 => (n-1)n(n+1) = 1*2*3 => Not a multiple of 4.
n=3 => (n-1)n(n+1)=2*3*4 =>Multiple of 4.

Hence not sufficient.

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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New post 19 Jan 2017, 00:05
For statement (2), everyone above seems to have plugged the values.

Is there a way to do it algebrically?

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Re: If n is a positive integer, is n^3 n divisible by 4 ?   [#permalink] 19 Jan 2017, 00:05
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If n is a positive integer, is n^3 n divisible by 4 ?

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