If n is an integer, which of the following must be divisible

[BTGmoderatorLU]

If n is an integer, which of the following must be divisible by 3?

$$A.\ n^3-4n$$
$$B.\ n^3+4n$$
$$C.\ n^2+1$$
$$D.\ n^2-1$$
$$E.\ n^2-4$$

The OA is A.

Please, can someone assist me with this PS question? I'm not sure about how can I solve it. Thanks in advance.

[Shahrukh@mbabreakspace]

Since we want to check divisibility by 3, Let us divide all the integers into 3 types: 3k, 3k+1 and 3k+2
If n is of 3k form:
then if n^3-4n can be written as n*(n^2-4)
So, for n=3k, it will become 3k*(9K^2-4), which is clearly divisible by 3
Now for 3k+1:
n*(n^2-4) will become (3k+1)*(9k^2+6k+1-4)= (3k+1)*(9k^2+6k- 3)= 3*(3k+1)*(3k^2+2k- 1), which is clearly divisible by 3
Now for 3k+2:
n*(n^2-4) will become (3k+2)*(9k^2+12k+4-4)= (3k+2)*(9k^2+12k)= 3*(3k+2)*(3k^2+4k), which is clearly divisible by 3

So, clearly, Option A is correct

[Scott@TargetTestPrep]

If n is an integer, which of the following must be divisible by 3?

$$A.\ n^3-4n$$
$$B.\ n^3+4n$$
$$C.\ n^2+1$$
$$D.\ n^2-1$$
$$E.\ n^2-4$$

Let's analyze each answer choice:

A. n^3 - 4n

n^3 - 4n = n(n^2 - 4) = n(n - 2)(n + 2) = (n - 2)n(n + 2)

We see that this is a product of 3 consecutive odd integers (if n is odd) or a product of 3 consecutive even integers (if n is even). Either way, at least one of the factors is a multiple of 3, thus the product is divisible by 3.

Let's illustrate this with some numbers. If we let n = 8, then n - 2 = 6 and n + 2 = 10, and we see that 6 is evenly divisible by 3. If we let n = 7, then n - 2 = 5 and n + 2 = 9, and we see that 9 is evenly divisible by 3. No matter what value we choose for n, we see that one of the values (n or (n - 2) or (n + 2)) will always be divisible by 3.

Answer: A