which parameters always make this rational equation evenly divisible? Hi guys I have the following equation:
$$x = \dfrac{a + b \times c - b}{c}$$
This is what I know about each variable:
$$a \ge 64$$
$$b \ge 0$$
$$8 \le c \le a$$
My questions is there a concise way for me to pick a, b, and c so that x will always be a positive integer >= 8. 
For example if $a = 64$; $b= 0$; $c = 8$ (the smallest possible values for each)
$x = (64 + 0 \times 8 - 0)/8$
$x = 8$ which is a positive integer $\ge 8$
However, if let's say $a = 200$; $b = 2$; $c = 10$
$x = (200 + 2 \times 10 - 2)/10$
$x = \dfrac{218}{10}$ which is $21.8$ and is not a positive integer greater than or equal to $ 8$
How can I always know that $x$ will be a positive integer?? Thanks!

As $$x = \frac{a+ b\cdot c -b}{c} = \frac{a-b}{c} + \frac{b\cdot c}{c} = \frac{a-b}{c} + b,$$ we can say that $x$ will be an integer whenever $a-b$ is divisible by $c$. That is, we need an integer $n$ such that $$a-b=cn \leftrightarrow a = cn + b.$$ Finally, as we want $x \geq 8$ this means we need $$\frac{a-b}{c} + b \geq 8.$$ By solving for $a$, we have $a \geq b -bc +8c$.
Combining our expressions involving $a$ results in $$cn+b \geq b-bc+8c$$ $$n+b\geq 8.$$ Now you may choose any $b \geq 0$ and any $n$ satisfying $n+b \geq 8$ and you will have $x \geq 8$.
