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Bunuel
Q = x^3 − x

Given that x is a positive integer such that x ≥ 75, which of the following is the remainder when Q is divided by 6?

A. 0
B. 1
C. 3
D. 5
E. Cannot be determined by the information provided


The answer does not depend on value of x.

\(x^3-x=x(x^2-1)=x(x-1)(x+1)\), where x is an integer.
So, (x-1), x and (x+1) are consecutive integers.

The product of 3 consecutive integers will have at least one even number and one multiple of 3.
That is, the product of 3 consecutive integers will always be divisible by 2*3 or 6.


A
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Bunuel
Q = x^3 − x

Given that x is a positive integer such that x ≥ 75, which of the following is the remainder when Q is divided by 6?

A. 0
B. 1
C. 3
D. 5
E. Cannot be determined by the information provided

We can rewrite the expression x^3 - x as x(x^2 - 1) = x(x - 1)(x + 1) = (x - 1)(x)(x + 1). Thus, Q is a product of three consecutive integers. Since the product of n consecutive integers is always divisible by n!, the product of three consecutive integers is divisible by 3! = 6, which means the remainder when Q is divided by 6 is 0.

Answer: A
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