Let $a_{0}$ be a positive integer and $a_{n}=5 a_{n-1}+4$, for all $n \geq 1$. Can $a_{0}$ be chosen so that $a_{54}$ is a multiple of 2013?
Let $x_{n}=\frac{a_{n}}{5^{n}}$. Then $x_{0}=a$ and $5^{n} x_{n}=a_{n}=5 a_{n-1}+4=5^{n} x_{n-1}+4$. So $x_{n}=x_{n-1}+\frac{4}{5^{n}}$. By induction,

$$
x_{n}=x_{0}+\left(\frac{4}{5}+\frac{4}{5^{2}}+\cdots+\frac{4}{5^{n}}\right)=a+\frac{4}{5}\left(1+\frac{1}{5}+\cdots+\frac{1}{5^{n-1}}\right)=a+\frac{4}{5} \cdot \frac{1-\frac{1}{5^{n}}}{1-\frac{1}{5}}=a+1-\frac{1}{5^{n}}
$$

So $a_{n}=5^{n} x_{n}=5^{n}(a+1)-1$. Now 2013 and $5^{n}$ are relatively prime. So there is a $b, 0<b<2013$, also relatively prime to 2013, such that $5^{54}=2013 c+b$. To have 2013 as a factor of $a_{54}$, it suffices to find an integer $y$ such that $(a+1) b-1=2013 y$. But this is a linear Diophantine equation in $a+1$ and $y$; it has an infinite family of solutions, among them such that $a+1 \geq 2$.