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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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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.

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Math Expert V
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Re: Is n divisible by 8?  [#permalink]

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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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Math Expert V
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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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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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2
1
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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2
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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1
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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2
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.

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

Please read the whole thread: if-x-3-x-n-and-x-is-a-positive-integer-greater-than-1-is-n-126854.html#p1207542

(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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GRE 1: Q169 V154 Re: If x^3-x=n and x is a positive integer greater than 1, is n  [#permalink]

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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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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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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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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

Answer: 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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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.

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

x3-x =n
& x(2,3,...) Is x divisible by 8 ?

stmt1: 3x = 2q + 1
=> x = odd and can be represented by 2k+1

x3-x = x(x+1)(x-1)
=(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 x3-x (=n) is divisible by8 => Sufficient

Stmt 2: x=4y+1
n= x(x+1)(x-1) = (4y+1)(4y+2)(4y)
8(4y+1)(2y+1)(y)
=> n is divisbile by 8 => Sufficient

Both stmts are sufficient by itself
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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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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.

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?
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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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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^3-x=n and x is a positive integer greater than 1, is n   [#permalink] 11 Apr 2019, 07:10
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