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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, 15:25
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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, 15:34
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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, 07:46
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.

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
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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, 07:49
Samirc2 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.

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

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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, 08:31
Bunuel wrote:
Samirc2 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.

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.
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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, 08:33
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, 09:08
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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, 05:48
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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, 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 (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, 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:
[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, 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?
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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, 08:10
(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, 08:14
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, 09:18
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, 22:47
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, 10:45
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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09 Jul 2015, 23:27
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.

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.

(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.
How did u deduce this -
x=even+odd=odd --> (x-1)x(x+1)
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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 2015, 00:47
honchos 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.

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.

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

Have you read the highlighted part?
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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 Aug 2016, 13:58
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on the right it says work/rate problems, I assume this is a mistake and it is divisibility / fractions / decimals?
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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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22 Aug 2016, 00:19
clarkie wrote:
on the right it says work/rate problems, I assume this is a mistake and it is divisibility / fractions / decimals?

Edited the tags. Than you.
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Re: If x^3-x=n and x is a positive integer greater than 1, is n   [#permalink] 22 Aug 2016, 00:19
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