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DS divisiblity question

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DS divisiblity question

by vinni.k » Tue Oct 23, 2012 11:53 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.

Answer is D

I got this question quickly, but i got confused with statement 1. (1). says there is a remainder . What does a remainder means here ?
I am taking "a remainder" as 0 and any number that comes as a remainder after division.
So, i selected B as the answer.

I am bit confused with "a remainder" ?

Looking forward to all you replies.
An expert reply will be really appreciated.

Regards
Vinni
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by anuprajan5 » Wed Oct 24, 2012 12:16 am
Hi Vinni,

Understand this from the perspective of division.

If you divide 7 apples amongst 2 people equally, then there is a remainder of 1 apple.
If you divide 6 apples amongst 2 people equally, there is nothing remaining and hence no remainder.

the first statement basically translates to the fact that x is odd and if x is odd, then x-1 and x+1, the 2 other divisors of n are even. In anny set of 3 consecutive integers, you will find one even number that is divisble by 2 and the 2nd even number divisble by 4.
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by vinni.k » Wed Oct 24, 2012 12:43 am
anuprajan5 wrote:
If you divide 7 apples amongst 2 people equally, then there is a remainder of 1 apple.
If you divide 6 apples amongst 2 people equally, there is nothing remaining and hence no remainder.
Thanks Anup.
So, 0 is no remainder.

Regards
Vinni
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by Brent@GMATPrepNow » Wed Oct 24, 2012 8:18 am
vinni.k 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.
Target question: Is n divisible by 8?

First notice that x^3 - x = x(x^2 - 1) = x(x+1)(x-1) = (x-1)(x)(x+1)
Also notice that x-1, x and x+1 represent 3 consecutive integers.
So, the question is telling (in a tricky way) that n equals the product of 3 consecutive integers.

BIG question: Under what conditions will the product of 3 consecutive integers be divisible by 8?
Well, this will occur if one of the 3 integers is divisible by 8 or if the middle number (x) is odd.
Why is this?
If the middle number is odd, then the numbers before and after will be even. Since every second even number is divisible by 4, we get a product in which one number is divisible by 2, one number is odd and the other number is divisible by 4.
If one number is divisible by 2, and another number is divisible by 4, the product must be divisible by 8

Okay, onto the statements.

Statement 1: When 3x is divided by 2, there is a remainder.
When any number is divided by 2, the remainder is either 0 or 1.
If that number is even, the remainder is 0.
If that number is odd, the remainder is 1.
So, statement 1 is telling us that 3x is odd.
This means that x must be odd.
If x (the middle number) is odd, then (according to my point in green above) n must be divisible by 8
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x = 4y + 1, where y is an integer.
First, 4y will be even for any integer value of y
So, 4y+1 must be odd
So, x must be odd
If x (the middle number) is odd, then (according to my point in green above) n must be divisible by 8
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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