Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 438
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
31 Jan 2012, 15:25
Question Stats:
63% (02:09) correct 37% (02:32) wrong based on 958 sessions
HideShow timer Statistics
If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730




Math Expert
Joined: 02 Sep 2009
Posts: 65764

Re: Is n divisible by 8?
[#permalink]
Show Tags
31 Jan 2012, 15:34
enigma123 wrote: If \(x^3\) – x = n and x is a positive integer greater than 1, is n divisible by 8?
(1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer.
I will really appreciate if you can tell me whether I am right or wrong. I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question. Yes, your reasoning is correct. If x^3  x = n and x is a positive integer greater than 1, is n divisible by 8? x^3x=x(x^21)=(x1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8 (basically if x=odd then (x1)x(x+1) will be divisible by 8*3=24). (1) When 3x is divided by 2, there is a remainder > 3x=odd > x=odd > (x1)x(x+1) is divisible by 8. Sufficient. (2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient. Answer: D. Hope it helps.
_________________




Intern
Joined: 11 Nov 2013
Posts: 15
GMAT Date: 12262013
GPA: 3.6
WE: Consulting (Computer Software)

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
04 Dec 2013, 11:47
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question. Statement 1: 3x/2 gives remainder . This means x is odd. odd^3 = odd and odd odd = even. x >1 and an odd integer . Lets take x = 3 n = x^3  x = 273 n= 24. 24/8 >Remainder = 0 Lets take x = 5 n = x^3  x = 1255 n= 120. 120/8 >Remainder = 0 Sufficient Statement 2: x = 4y+1 . y>integer Put x in equation n = x^3  x n = (4y+1)^3  (4y+1) apply formula(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 (4y+1)^3 = 64y^3 + 3*16x^2 + 3*4y*1 + 1 = 64y^3 + 48x^2 + 12y + 1 n = 64y^3 + 48x^2 + 12y + 1  4y  1 n = 64y^3 + 48x^2 + 8y n = 8(8y^3 + 6x^2+1) Hence n is divisible by 8. > Sufficient Hence D




Math Expert
Joined: 02 Sep 2009
Posts: 65764

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
03 Apr 2013, 08:33
Samirc2 wrote: Bunuel wrote: Samirc2 wrote: Hi,
Quick question: I do not understand why, when considering the product (x1)x(x+1), you conclude that one of the factor is divisible by 4.
Let me explain my line of reasoning here: we know x is odd, so both (x1) and (x+1) are even, hence my conclusion is that (x1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x1)x(x+1) is divisible by 4 and not by 8. Could you please clarify this point?
Thanks
By the way awesome book, thanks for sharing
Sam
Please check here: ifx3xnandxisapositiveintegergreaterthan1isn126854.html#p1037233As you could see from my post, I did check the previous posts and quoted them but I cannot figure out what is wrong with my reasoning. (x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8
_________________



Manager
Status: Looking to improve
Joined: 15 Jan 2013
Posts: 142
GMAT 1: 530 Q43 V20 GMAT 2: 560 Q42 V25 GMAT 3: 650 Q48 V31

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
03 Apr 2013, 09:08
Just list out all the even numbers  0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0
so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.
Hope this helps..



Intern
Joined: 23 Oct 2012
Posts: 21

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
12 Oct 2013, 05:48
nt2010 wrote: Just list out all the even numbers  0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0
so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.
Hope this helps.. I believe the last statement would be correct only if the sequence starts with even integer 2. For instance, if there are two consecutive even numbers, 0 and 2, then 4 is not a factor. However, if the two consecutive numbers are 2 and 4, then 4 is a factor of 2*4.



Math Expert
Joined: 02 Sep 2009
Posts: 65764

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
12 Oct 2013, 08:06
audiogal101 wrote: nt2010 wrote: Just list out all the even numbers  0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0
so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.
Hope this helps.. I believe the last statement would be correct only if the sequence starts with even integer 2. For instance, if there are two consecutive even numbers, 0 and 2, then 4 is not a factor. However, if the two consecutive numbers are 2 and 4, then 4 is a factor of 2*4. That's not true. 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). Thus if (x1)x(x+1)=0*1*2=0, then the product is still divisible by 8. Hope it's clear.
_________________



Manager
Joined: 11 Sep 2012
Posts: 82

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
25 Dec 2013, 11:56
Is there a general pattern for what the remainder is when the square of an odd number is divisible by even numbers?
For instance, is the remainder always 1 when divided by 2, 4 and 8?



Manager
Joined: 28 Dec 2013
Posts: 65

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
20 Jun 2014, 08:10
(2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient.
Question : x = even + odd + odd, where does this come from? I think it comes from (2) but I don't get how if that's so.



Math Expert
Joined: 02 Sep 2009
Posts: 65764

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
20 Jun 2014, 08:14
sagnik242 wrote: (2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient.
Question : x = even + odd + odd, where does this come from? I think it comes from (2) but I don't get how if that's so. \(x = 4y + 1\). Now, \(4y\) is even, because of 4, and 1 is odd, thus \(x=even+odd=odd\). Does this make sense?
_________________



Manager
Joined: 28 Dec 2013
Posts: 65

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
20 Jun 2014, 09:18
Bunuel wrote: sagnik242 wrote: (2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient.
Question : x = even + odd + odd, where does this come from? I think it comes from (2) but I don't get how if that's so. \(x = 4y + 1\). Now, \(4y\) is even, because of 4, and 1 is odd, thus \(x=even+odd=odd\). Does this make sense? Yes thanks so much:)



Intern
Joined: 14 Feb 2013
Posts: 19

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
10 Jul 2014, 22:47
if x is odd, how is x(x+1)(x1) divisible by 8. I can't understand. Help?



Math Expert
Joined: 02 Sep 2009
Posts: 65764

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
11 Jul 2014, 10:45
hamzakb wrote: if x is odd, how is x(x+1)(x1) divisible by 8. I can't understand. Help? Please read the whole thread: ifx3xnandxisapositiveintegergreaterthan1isn126854.html#p1207542(x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8
_________________



Current Student
Joined: 12 Aug 2015
Posts: 2516

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
14 Jan 2017, 04:43
Great Question. Here is what i did in this one=> As n=x^3x=> (x1)*x*(x+1)=> If x is odd => both x1 and x+1 will be consecutive even integers.So one of them would be multiple of 4 and other would be a multiple of 2 for sure.Hence n would be a multiple of 8.
Statement 1> 3x is odd => x is odd => Sufficient. Statement 2> x=4y+1 => Even+odd=odd Hence Sufficient.
Hence D
_________________



Intern
Joined: 18 Sep 2017
Posts: 1

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
21 Sep 2017, 02:23
Hi, I don't understand why if x is odd then (x1)x(x+1) will be divisible by 8*3=24. Can someone pls help?



Math Expert
Joined: 02 Sep 2009
Posts: 65764

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
11 Oct 2017, 20:24
JsZJ wrote: Hi, I don't understand why if x is odd then (x1)x(x+1) will be divisible by 8*3=24. Can someone pls help? x^3x=x(x^21)=(x1)x(x+1), notice that we have the product of three consecutive integers, so one of (x1), x or (x+1) is divisible by 3. Now, if x=odd, then (x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8 (basically if x=odd then (x1)x(x+1) will be divisible by 8*3=24).
_________________



RSM Erasmus Moderator
Joined: 26 Mar 2013
Posts: 2493
Concentration: Operations, Strategy

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
12 Oct 2017, 01:42
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8?
(1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer.
x & n are Positive integers, so zero is not considered here and x > 1 n = x^3 – x = x (x^2  1) 1) 3x/2 yields reminder.........means x is odd number Let x = 3 ..........3 (91) = 3 * 8...........Answer is yes Let x = 5 ..........5 (251) = 5 *24...........Answer is yes Let x = 7 ..........7 (491) = 7 *4 8...........Answer is yes Let x = 9 ..........9 (811) = 9 * 80...........Answer is yes Let x = 15 ..........15 (2251) = 15 * 224...........Answer is yes There is pattern here Sufficient (2) x = 4y + 1, where y is an integer. Here, no need to rush in plugging number...........the equation yields odd number as above Sufficient Answer: D



Manager
Joined: 23 Oct 2017
Posts: 61

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
03 Jan 2018, 18:34
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question.  x3x =n & x(2,3,...) Is x divisible by 8 ? stmt1: 3x = 2q + 1 => x = odd and can be represented by 2k+1 x3x = x(x+1)(x1) =(2k+1)(2k+2)(2k) =4(2k+1)(k+1)(k) Now (k+1)(k) is a product of 2 consecutive integers, one of which HAS to be even thus this product is even so x3x (=n) is divisible by8 => Sufficient Stmt 2: x=4y+1 n= x(x+1)(x1) = (4y+1)(4y+2)(4y) 8(4y+1)(2y+1)(y) => n is divisbile by 8 => Sufficient Both stmts are sufficient by itself



Manager
Joined: 29 May 2017
Posts: 188
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
10 Sep 2018, 18:13
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question. my 'equal' key is not working so using colon : as equal!! 3x:2p+2 p:2 > 3x:6 >x:2...... 1x2x3 ....not div by 8 p:3 > 3x:8 >x:8/3..... 5/3 x 8/3 x 11/3 .... div by 8 p:4 > 3x:10 > x:10/3 .... 7/3 x 10/3 x 13/3 .... not div by 8 then how come 1 is sufficient?



Intern
Joined: 10 Apr 2019
Posts: 2

Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
Show Tags
11 Apr 2019, 06:10
Good question.
Stmt 1: 3x=2a+r ===> x=(2a+r)/3. Since x is an integer, then x is divisible by 3 and minimum x=3 ===> (2)*(3)*(4) is divisible by 8 (and any other sequence is). Stmt 2: x=4y+1. Since x cant be 1 then y>0. For minimum y=1, then x=5 ===> (4)*(5)*(6) is divisible by 8 (and any other sequence is).
Therefore (D).




Re: If x^3x=n and x is a positive integer greater than 1, is n
[#permalink]
11 Apr 2019, 06:10



Go to page
1 2
Next
[ 23 posts ]

