x is an integer. If x>9, and x^3+3x^2+2x is not divisible

Solution #1: Algebraic

Key concept: When we list consecutive integers, every 4th value is a multiple of 4
{...-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,...}
So, among any 4 consecutive integers, there will be ONE value that is divisible by 4.

-----------ONTO THE QUESTION------------------------

Take \(x^3+3x^2+2x\)
Factor out the x to get: \(x(x^2+3x+2)\)
Factor the quadratic to get: \(x(x+1)(x+2)\)
Notice that x, (x+1), and (x+2) are CONSECUTIVE integers.
Since we're told that \(x^3+3x^2+2x\) is not divisible by 4, we know that none of the 3 values, x, (x+1), or (x+2), are divisible by 4.
So, among the FOUR consecutive integers (x-1), (x), (x+1), and (x+2), we know that x-1 must be divisible by 4 (since none of the other 3 values are divisible by 4)
In fact, if we list more consecutive integers in this form, we'll see that every FOURTH number will be divisible by 4.

We get: {...., (x-9), (x-8), (x-7), (x-6), (x-5), (x-4), (x-3), (x-2), (x-1), (x), (x+1), (x+2), (x+3),...}

So, we can see that (x-5) must be divisible by 4

Answer: C

Cheers,
Brent