which parameters always make this rational equation evenly divisible?

Question

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!

Answer 1

If $$x=(a+b*c-b)/c$$ You can split it into $x=\frac{a-b}{c}+b$

Then, the only divisibility condition is $$\frac{a-b}{c}=k$$ $$a-b=ck$$ $$a=b+ck$$ So, you can pick any $b$,$c$ and $k$ integer such that x is an integer.

Replacing a you get $k+b \ge 8$

Finally, you can pick any integers $b$ and $k$ such that $k+b \ge 8$ and any integer $c$ that follow your initial inequalities, then operate for $a$.

Answer 2

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 $x \geq 8$.