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If x^3-x=n and x is a positive integer greater than 1, is n

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enigma123
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If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   31 Jan 2012, 16:25
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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.

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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 (x-1), 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.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   31 Jan 2012, 16:34
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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 (x-1), 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^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)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 --> (x-1)x(x+1) is divisible by 8. Sufficient.

(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.

Answer: D.

Hope it helps.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   04 Dec 2013, 12:47
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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:
Show Spoiler
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 (x-1), 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 = 27-3
n= 24. 24/8 ->Remainder = 0

Lets take x = 5
n = x^3 - x = 125-5
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
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   03 Apr 2013, 09:33
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Samirc2 wrote:
Bunuel wrote:
Samirc2 wrote:

Hi,

Quick question: I do not understand why, when considering the product (x-1)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 (x-1) and (x+1) are even, hence my conclusion is that (x-1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x-1)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: if-x-3-x-n-and-x-is-a-positive-integer-greater-than-1-is-n-126854.html#p1037233


As 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.


(x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   03 Apr 2013, 10:08
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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..
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   12 Oct 2013, 06:48
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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.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   12 Oct 2013, 09:06
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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).

Thus if (x-1)x(x+1)=0*1*2=0, then the product is still divisible by 8.

Hope it's clear.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   20 Jun 2014, 09:10
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)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.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   20 Jun 2014, 09:14
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sagnik242 wrote:
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)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?
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   21 Sep 2017, 03:23
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Hi, I don't understand why if x is odd then (x-1)x(x+1) will be divisible by 8*3=24. Can someone pls help?
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   11 Oct 2017, 21:24
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JsZJ wrote:
Hi, I don't understand why if x is odd then (x-1)x(x+1) will be divisible by 8*3=24. Can someone pls help?


x^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers, so one of (x-1), x or (x+1) is divisible by 3. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)x(x+1) will be divisible by 8*3=24).
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   28 Mar 2020, 01:16
Bunuel wrote:
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 (x-1), 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^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)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 --> (x-1)x(x+1) is divisible by 8. Sufficient.

(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.

Answer: D.

Hope it helps.


-------------------------------------------

Here question states "x is a positive integer greater than 1" and in explanation we have focused only on "if x=odd" .I want to why we have not considered "x=even" case.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   28 Mar 2020, 08:13
x^3-x = n
x(x^2-1) = n

(x-1)(x)(x+1) =n

n is the product of 3 consecutive numbers.

There are 2 cases when product of 3 consecutive numbers is divisible by 8.

Case 1- when x is odd
Since x is odd, x-1 and x+1 must be even. Also, x-1 and x+1 are 2 consecutive even numbers, so one must be odd multiple of 2 and other is even multiple of 2. Hence, n is always divisible by 8, if x is odd.

Case 2- when x is a multiple of 8.
(As x-1 and x+1 both are odd when x is even, x must be multiple of 8.)

Both statements tells you that x is odd. Hence, Sufficient.




himanigahlot wrote:
Bunuel wrote:
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 (x-1), 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^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)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 --> (x-1)x(x+1) is divisible by 8. Sufficient.

(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.

Answer: D.

Hope it helps.


-------------------------------------------

Here question states "x is a positive integer greater than 1" and in explanation we have focused only on "if x=odd" .I want to why we have not considered "x=even" case.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   21 Nov 2020, 04:32
Bunuel
I tried to prove option A ;I understood the logical approach but the mathematical approach is showing a different result
as
x*(x+1)*(x+2)
1)--------------gives us x is odd therefore x is in form of 2k+1 where k->1,2,3,4.....
so substituting and solving
we get 8k^3+12k^2+4k now this term when divided by 8 will leave no remainder only when k is in (2,4,6,8... i.e even)
Hence mathematically its telling me its not necessarily divisble by 8 Can someone help?
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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink] New post   22 Nov 2020, 00:47
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DanTe02 wrote:
Bunuel
I tried to prove option A ;I understood the logical approach but the mathematical approach is showing a different result
as
x*(x+1)*(x+2)
1)--------------gives us x is odd therefore x is in form of 2k+1 where k->1,2,3,4.....
so substituting and solving
we get 8k^3+12k^2+4k now this term when divided by 8 will leave no remainder only when k is in (2,4,6,8... i.e even)
Hence mathematically its telling me its not necessarily divisble by 8 Can someone help?


Hey a couple of things here. The original expression is (x-1)(x)(x+1). Substituting, x=2k+1 in the original expression.
=> (2k)(2k+1)(2k+2).
=> 2(k)(2k+1)(2)(k+1)
=> 4(k^2+k) (2k+1).
Now, note that you already have 4 in the expression and k^2 + k will always be even, irrespective of whether k is odd or even. You can also try and put k=odd or even at any step and you will ge the same answer.

Hope this helps!
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