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

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

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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.

I will really appreciate if you can tell me whether I am right or wrong:
[Reveal] 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.
[Reveal] Spoiler: OA

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Re: Is n divisible by 8? [#permalink]

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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.

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.

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]

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03 Apr 2013, 09:33
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

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]

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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]

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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]

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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 except zero itself).

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]

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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:
[Reveal] 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]

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25 Dec 2013, 12:56
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?

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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]

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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]

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20 Jun 2014, 09:14
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]

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20 Jun 2014, 10:18
Bunuel wrote:
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?

Yes thanks so much:)

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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]

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10 Jul 2014, 23:47
if x is odd, how is x(x+1)(x-1) divisible by 8. I can't understand. Help?

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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]

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11 Jul 2014, 11:45
hamzakb wrote:
if x is odd, how is x(x+1)(x-1) divisible by 8. I can't understand. Help?

(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]

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14 Jan 2017, 05:43
Great Question.
Here is what i did in this one=>
As n=x^3-x=> (x-1)*x*(x+1)=> If x is odd => both x-1 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

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Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]

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21 Sep 2017, 03:23
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]

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11 Oct 2017, 21:24
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]

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12 Oct 2017, 02:42
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 (9-1) = 3 * 8...........Answer is yes

Let x = 5 ..........5 (25-1) = 5 *24...........Answer is yes

Let x = 7 ..........7 (49-1) = 7 *4 8...........Answer is yes

Let x = 9 ..........9 (81-1) = 9 * 80...........Answer is yes

Let x = 15 ..........15 (225-1) = 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

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Re: If x^3-x=n and x is a positive integer greater than 1, is n   [#permalink] 12 Oct 2017, 02:42
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