enigma123 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.
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.
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.
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.