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Divisibility question

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Divisibility question

by crackthetest » Sun Mar 07, 2010 8:44 pm
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
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by firdaus117 » Sun Mar 07, 2010 10:04 pm
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
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by kstv » Mon Mar 08, 2010 5:41 am
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.
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by firdaus117 » Mon Mar 08, 2010 8:05 am
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.
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by gmatmachoman » Mon Mar 08, 2010 8:12 am
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
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by kstv » Mon Mar 08, 2010 9:19 am
Just proves that the problems are difficult cos' of the time factor.
firdaus117 is right. its D. But no 2nd chance.
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by Anaira Mitch » Wed Dec 28, 2016 6:52 am
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.
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