Similarly, we can replace 3000 with 3, since \(1000 = 1 + 999 \equiv 1 \pmod 9\text{.}\) So our original congruence becomes

\begin{equation*} x \equiv 3 + 4 + 0 + 1 \pmod 9 \end{equation*}

\begin{equation*} x \equiv 8 \pmod 9\text{.} \end{equation*}

Therefore \(3491\) divided by 9 has remainder 8.