Infinitely many positive integers of the form $1998k+1$ such that all digits in their decimal representation are equal 
Prove that there exist infinitely many positive integers of the form $1998k+1, k \in \mathbb{N},$ such that all digits in their decimal representation are equal. 

We need to find integer solutions to $a \cdot \dfrac{10^n-1}{9} = 1998k+1$. Thus $$a(10^n-1) = 9(1998k+1).$$ How can we find infinitely many integer solutions to this?

Credit of the full answer to Mlazhinka Shung Gronzalez LeWy.
Let $N = 10^{3n} - 1 = 999\ldots 9$. By factorisation, $N$ is a multiple of $999$, and the other quotient is $$\frac{10^{3n}-1}{10^3-1} = 10^{3(n-1)}+ \cdots + 10^6 + 10^3+1.$$
I further would like the quotient be a multiple of $9$, so that $N$ is a multiple of $999\times 9$. Since $10\equiv 1\pmod 9$, the right hand side
$$10^{3(n-1)}+ \cdots + 10^6 + 10^3+1 \equiv n \pmod 9$$
So $n$ can be any multiple of $9$, i.e. $N = 10^{27m}-1$. Let 
$$\begin{align*}
M = \frac N9 &= \frac{10^{27m}-1}{9}\\
10M &=\frac{10^{27m+1}-1}{9}-1
\end{align*}$$
$M$ is a multiple of $999$, and so $10M$ is a multiple of $999$, and so 
$$\frac{10^{27m+1}-1}{9}\equiv 1 \pmod{999}$$
But the left hand side is just $10^{27m} + 10^{27m-1} + \cdots + 10^2+10^1+10^0$, which is odd, 
$$\frac{10^{27m+1}-1}{9}\equiv 1 \pmod{2}$$
By Chinese remainder theorem,
$$\frac{10^{27m+1}-1}{9}\equiv 1 \pmod{1998}$$
