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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 ? (1) n = 2k + 1, where k is an integer. (2) n^2 + n is divisible by 6.
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Originally posted by Baten80 on 23 Feb 2011, 22:25.
Last edited by Baten80 on 24 Feb 2011, 03:52, edited 1 time in total.




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Re: OG DS 170Arithmetic
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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 ?
(1) n = 2k + 1, where k is an integer
(2) n^2 + n is divisible by 6
Originally posted by Impenetrable on 12 Mar 2012, 02:52.
Last edited by Bunuel on 20 Nov 2014, 07:53, edited 2 times in total.
Edited the question and the OA



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Re: If n is a positive integer, is n^3 – n divisible by 4 ?
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12 Mar 2012, 03:13
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^3n=n(n^21)=(n1)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 n1 and n+1 are even hence (n1)n(n+1) is divisible by 4. Sufficient. (2) n^2 + n is divisible by 6 > if n=2 then n^3n=6 and the answer is NO but if n=3 then n^3n=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 ?
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20 Mar 2012, 19:00
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^3n 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^3n is div by 4
Let's try: n = 2 it satisfies n^2 + n (6) is div by 6; BUT n^3n(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^3n(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 ?
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20 Mar 2012, 22:52
From Statement I, n = 2k+1, therefore n is odd. n^3n = 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^3n 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 ?
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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.



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



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Re: If n is a positive integer, is n^3 n divisible by 4 ? (1) n
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18 Apr 2012, 04:14
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: mathnumbertheory88376.htmlHope it helps.
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If n is a positive integer, is n^3 n divisible by 4 ?
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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(91) 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 coprime 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
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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). Check Number Theory chapter of Math Book for more: mathnumbertheory88376.htmlHope it helps. Sorry Bunuel, just to make this clear if k = 0 then n = 1 (2 * 0 + 1) this means that (n1)(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
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15 Jun 2012, 19:02
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: mathnumbertheory88376.htmlHope it helps. Sorry Bunuel, just to make this clear if k = 0 then n = 1 (2 * 0 + 1) this means that (n1)(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 (n1)(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 ?
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15 Aug 2013, 21:14
X= n^3  n = n(n^21) = (n1)*n*(n+1)
REM(X/4)=?
(1).
n is odd => For every odd number there exists two even numbers in the expression (n1) 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 ?
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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 ? (n1)(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 ?
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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)(n1) 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 (n1) 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 ?
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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 ?
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15 Jan 2017, 01:01
Excellent Official Question. Here is what i did in this one =>
Firstly n^3n= (n1)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 => n1 will be even and n+1 will be even too. Further more one out of them will be multiple of 4. Thus (n1)n(n+1) will be multiple of 8. As it is already a multiple of 6 => we can conclude that (n1)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 (n1)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 => (n1)n(n+1) = 1*2*3 => Not a multiple of 4. n=3 => (n1)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 ?
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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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