brokerbevo wrote:
I'm having trouble with the rule behind this concept. This problem is a good example of my confusion:
If y is an integer, is y divisible by 3?
(1) y = 2x^3 + 9x^2 - 5x
(2) x is indivisible by 3
Please explain your rationale behind your answer.
Here is what I did:
(1) factor out an x --> x (2x^2 + 9x - 5)
factor the quadratic --> x (2x - 1) (x + 5)
so, y is the product of 3 integers and I wrote down that since 3 integers are multiplied together, it must be that it is divisible by 3. HOWEVER, I thought that this rule ONLY applies if the integers that are multiplied together are consecutive!!!
For example, x (x - 1) (x -4) is not divisible by 3 (even though it is the product of 3 numbers) because the parts are not consecutive.
Please explain...
Well, it doesn't *only* apply when the integers are consecutive. Say x is an integer. You know already that one of x, x+1 or x+2 is a multiple of 3. Well, that guarantees that one of, say, x, x+4, and x+8 is divisible by 3 (because if x+1 is, then x+1+3 = x+4 is, and if x+2 is, so is x+2 + 6 = x+8), and many other combinations besides.
You can apply that logic here if you want. We know that one of these numbers: x, x+1 or x+2 is a multiple of 3:
-If x is a multiple of 3, then x is a multiple of 3
-If x+1 is a multiple of 3, then 2x+2 is also a multiple of 3, and (subtract 3) 2x-1 is also a multiple of 3
-If x+2 is a multiple of 3, then x+2+3 = x+5 is also a multiple of 3.
So one of x, 2x-1 or x+5 will always be a multiple of 3, and since y = x (2x - 1) (x + 5), y must be a multiple of 3.
There are other ways to go about the question that might be preferable here (modular arithmetic, for example); I just wanted to show how you can extend the logic about 'consecutive numbers' to this situation.