Is there any way to solve this problem without having to do it by hand?

Question

I'm dealing with the following problem in computational programming. I'm trying to find a way to build an algorithm that can quickly resolve the following problem statement. Is there any way to group the relations below, or to find values of y in the following problem statement very efficiently without having to do it by hand?

Problem: Given an integer value x, where x > 2, is there any integer value y, where y >= 0, such that x satisfies any of these relations for some integer k, where k > 0?

[1]: y = (x - 2k)/(20k)
[2]: y = (x - 6k)/(20k)
[3]: y = (x - 14k)/(20k)
[4]: y = (x - 18k)/(20k)

Example #1: If x = 4:

From [1]: y = (4 - 2k)/(20k) = 0 for k = 2.

Example #2: If x = 6:

From [1]: y = (6 - 2k)/(20k) = 0 for k = 3.

Example #3: If x = 3:

From [1]: y = (3 - 2k)/(20k) ... k = 1 => y = (3 - 2(1))/(20(1)) = 1/20, which would not be an integer ... k = 2 => y = (3 - 2(2))/(20(2)), which would be negative. It would be negative for any k > 1.
From [2]: y = (3 - 6k)/(20k) ... It would be negative for any k > 0.
From [3]: y = (3 - 14k)/(20k) ... It would be negative for any k > 0.
From [4]: y = (3 - 18k)/(20k) ... It would be negative for any k > 0.

We can deduce that for x = 3, there is no value y that satisfies any of the above relations.

Answer 1

As stated, the problem has a very simple answer. If $x$ is even, relation [1] is satisfied if you let $y=0$ and $k=x/2$. If $x$ is odd, none of the relations can be satisfied, since they all imply that $x$ is even -- i.e., they can be rewritten as saying $x=2k(10y+r)$ with $r=1$, $3$, $7$, or $9$.

Answer 2

The most restrictive is [4], which tells you that $$ x = 2 (10 k y + 9) $$ So $x$ is even, and moreover $$ \frac{1}{10} \left( \frac{x}{2} - 9 \right) = k y $$ has to be an integer, i.e. for some integer $c$: $$ x = 20 c + 18 $$ This will comply with [4], and also [1] and [2]. [3] can be handled similarly, combining with this will give a way to comply with all four.