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Re: Is n divisible by 8? [#permalink]
31 Jan 2012, 15:34

2

This post received KUDOS

Expert's post

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.

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

Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
03 Apr 2013, 07:49

Expert's post

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?

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

Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
03 Apr 2013, 08:33

Expert's post

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?

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 _________________

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

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

Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
12 Oct 2013, 08:06

Expert's post

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.

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

(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 _________________