Mathematical way to solve integer numbers $217 = (20x+3)r+x$ Is there any mathematical way to find the integer numbers that solve the following equation:
$$217 = (20x+3)r+x$$

HINT:
$$\implies r=\frac{217-x}{20x+3}$$
If integer $d$ divides both the numerator & the denominator; it will divide
$\displaystyle(217-x)\cdot20+(20x+3)\cdot1=4343=43\cdot101$
So, $\displaystyle20x+3$ must divide $4343$ to keep $r$ an integer
Fortunately, the number of divisors are limited  namely, $\displaystyle\pm1,\pm4343,\pm43,\pm101$
Again $\displaystyle\pm1,\pm101\equiv\pm1\pmod{20};$ and $\displaystyle-43,-4343\equiv-3\pmod{20}$ unlike $\displaystyle20x+3\equiv3$ which leaves only two options, right?
