4. Arrange $1, 2, 3$ in a cyclic order from left to right, forming a 2009-digit number: $x=123123 \cdots 12312$. Find the remainder when $x$ is divided by 101.
(Tip: Since $100 \equiv -1 \pmod{101}$, we have
$$
\begin{array}{l}
123123 = 123 \times 100 + 123 \\
\equiv -123 + 123 \equiv 0 \pmod{101}.
\end{array}
$$

Therefore, $123123 \cdots 12312 = 123123 \cdots 12300 + 12$
$$
\equiv 0 + 12 \equiv 12 \pmod{101}.
$$
)