Hi Richachampion,
This is a YES-NO DS question.
Question: Is n divisible by 8 ?
Where n = x^3 - x,
Remember a number to be divisible by 8, it should have minimum of three 2's.
Now lets little bit analyze "n" here,
Given, x^3-x=x*(x^2-1)=(x-1)*x*(x+1),
Notice that we have the product of three consecutive integers (It's given that x is a positive integer)
Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4.
For example if x = 5 then 4 * 5 * 6 will have minimum three 2's
If x is odd then it is divisible by 8.
So we need check only whether x is odd.
Statement I is sufficient: When 3x is divided by 2, there is a remainder.
If x is even then 3x is divisible by 2, so the remainder is zero then which is not possible according to statement I.
So x has to be odd. So sufficient, where the remainder is one.
Statement II is sufficient: x = 4y + 1, where y is an integer.
Even + odd = Odd,
Here 4y is even and 1 is odd, so 4y+1 is always odd.
So x is odd.
So sufficient.
So the answer is D.
Here statement I is poorly worded (Don't know what's the source of the question) generally you don't see these type of official GMAC questions.
When 3x is divided by 2, there is a remainder - If we look at look it zero is also consider as a remainder, in that case x could be even as well. So it should be framed something like this "3x is not perfectly divisible by 2"
Hope this is clear.
