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

Question

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.

Bunuel · Answer

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.

Bunuel · Answer

(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

nt2010 · Answer

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..

audiogal101 · Answer

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.

Bunuel · Answer

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.

A4G · Answer

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

sagnik242 · Answer

(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.

Bunuel · Answer

x = 4y + 1 . Now,  4y  is even, because of 4, and 1 is odd, thus  x=even+odd=odd .

Does this make sense?

JsZJ · Answer

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?

Bunuel · Answer

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).

himanigahlot · Answer

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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.

nick1816 · Answer

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.

DanTe02 · Answer

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?

MBAB123 · Answer

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!

Saya12345 · Answer

x-1 x x+1

statement one gives x is odd 
implies x-1 x+1 are consecutive evens 
hence, one of then divisible by 2, other divisible by 4
Thus, the entire number is div by 8 
-> S1 is sufficient.

S2
x=4y+1
4y 4y+1 4y+1
div by 4 & 2
hence div by 8 
S2 is sufficient.

Hence, Ans = D

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

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