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We could do long division, but there is another way. We want to find \(x\) such that \(x \equiv 3491 \pmod{9}\text{.}\) Now \(3491 = 3000 + 400 + 90 + 1\text{.}\) Of course \(90 \equiv 0 \pmod 9\text{,}\) so we can replace the 90 in the sum with 0. Why is this okay? We are actually subtracting the “same” thing from both sides:

\begin{equation*} \begin{aligned}x \amp \equiv 3000 + 400 + 90 + 1 \pmod 9 \\ - ~~ 0 \amp \equiv 90 \pmod 9 \\ x \amp \equiv 3000 + 400 + 0 + 1\pmod 9. \end{aligned} \end{equation*}

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