Let's make this question a bit simpler:
It is given that n=x^3-x
=x(x^2-1)
=x(x-1)(x+1)
n =(x-1)x(x+1)
So,we can see that n is a product of three consecutive integers.
Statement 1: 3x is not divisible by 2.So,3x is an odd number which implies x is an odd number.
So,let us say x=2k+1 for k=1,2,3,4,5........
n= (2k+1-1)(2k+1)(2k+1+1)
= 2k *(2k+1)(2K+2)
=4*k(k+1)(2k+1)
Now, k(k+1) is a product of two consecutive integers.hence one of them is an even number which makes n a factor of 8.
So, n is divisible by 8. Sufficient.
Statement 2: x=4y+1
n= 4y(4y+1)(4y+2)
Similarly,(4y+1)(4y+2) is a product of two consecutive integers.hence one of them is an even number which makes n a factor of 8.
So, n is divisible by 8. Sufficient.
Option D

x³-x = n or x(x-1)(x+1) = n
1) 3x/2 has no remainders so x is atleast 2 or multiple of 2
2*1*3 is not divisible by 8 , infact n is divisible by 9 (Out of scope, pl ignore). Insuff
2) x=4y+1, x-1 = 4y , x+1 = 4y+2
n= x(4y)(4y+2) = 8xy(2y+1) . Suff
IMO B
Maybe I am wrong but in 2 min this is my answer.

kstv wrote:x³-x = n or x(x-1)(x+1) = n
1) 3x/2 has no remainders so x is atleast 2 or multiple of 2
2*1*3 is not divisible by 8 , infact n is divisible by 9 (Out of scope, pl ignore). Insuff
2) x=4y+1, x-1 = 4y , x+1 = 4y+2
n= x(4y)(4y+2) = 8xy(2y+1) . Suff
IMO B
Maybe I am wrong but in 2 min this is my answer.

Is it really what first statement is saying?
(1) When 3x is divided by 2, there is a remainder.

crackthetest 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.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.

OA - D

From 1 we can say X is a odd number. So start plugging odd values for X in (x^3 - x )

We will try some examples: Let x =5, then x^3 - x =120 which is divisible by 8

let X=7, x^3 - x =336 which is divisible by 8..

So st 1 is sufficient

Now coming to st 2:

x= 4y+1, X will be a odd number for any value of y

So again st 2is sufficient to say whetehr n is divisible by 8

IMO D , individually we can say whetehr n is divisible by 8 or not

Just proves that the problems are difficult cos' of the time factor.
firdaus117 is right. its D. But no 2nd chance.

If we factor the equation in the question, we get n = x(x - 1)(x + 1) or n = (x - 1)x(x +1). n is the product of three consecutive integers. What would it take for n to be divisible by 8? To be divisible by 8, is to be divisible by 2 three times, or to have three 2's in the prime box.
The easiest way for this to happen is if x is odd. If x is odd, both x - 1 and x + 1 will be even or divisible by 2. Furthermore, if x is odd, x - 1 and x + 1 will also be consecutive even integers. Among consecutive even integers, every other even integer is divisible not only by 2 but also by 4. Thus, either x - 1 or x + 1 must be divisible by 4. With one number divisible by 2 and the other by 4, the product represented by n will be divisible by 8 if x is odd.

(1) SUFFICIENT: This tells us that x is odd. If 3x divided by 2 has a remainder, 3x is odd. If 3x is odd, x must be odd as well.

(2) SUFFICIENT: This statement tells us that x divided by 4 has a remainder of 1. This also tells us that x is odd because an even number would have an even remainder when divided by 4. Alternative method: if we rewrite this statement as x - 1 = 4y, we see that x - 1 is divisible by 4, which means that x + 1 is also even and the product n is divisible by 8.
The correct answer is D.