Q: How can I solve this problem without doing it by hand? 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 without forcing me to do it by hand. Is there any way to group the following relations below by some pattern or identity in order to make finding a possible values of c much more efficiently?

Problem: Given any integer x, where x > 2, is there any integer c,
where c >= 1 that satisfies the following relations for some integer k,
where k >= 0.


*

*$\frac{x}{11 + 10k} = c \Rightarrow x = 11c + 10ck$

*$\frac{x}{3  + 10k} = c \Rightarrow x = 3c  + 10ck$

*$\frac{x}{7  + 10k} = c \Rightarrow x = 7c  + 10ck$

*$\frac{x}{9  + 10k} = c \Rightarrow x = 9c  + 10ck$
Example $\#1$: $x = 3$
From $\#2$: $3 = 3c + 10ck$, so $k = 0, c = 1$ satisfy this relation.
Example $\#2$: $x = 4$
From $\#1$: $c = \frac{4}{11 + 10k}$, so for any $k \ge 0$, the denominator will always be larger than $4$. No integer value of $c \ge 1$ will ever satisfy this equation.
From $\#2$: $c = \frac{4}{3  + 10k}$, so for $k = 0$, the denominator does not divide evenly into $4$. For $k \ge 1$, the denominator will always be larger than $4$. No integer value of $c \ge 1$ will ever satisfy this equation.
From $\#3$: $c =\frac{4}{7  + 10k}$, so for any $k \ge 0$, the denominator will always be larger than $4$. No integer value of $c \ge 1$ will ever satisfy this equation.
From $\#4$: $c = \frac4{9  + 10k}$, so for any $k \ge 0$, the denominator will always be larger than $4$. No integer value of $c \ge 1$ will ever satisfy this equation.

A: Your relations pick out the numbers $x$ that have an odd prime divisor not equal to $5$.  Hence the only numbers not covered are those of the form $x=2^m\cdot5^n$.
