Solution 5.2.3.1.

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*}

Next, note that \(400 = 4 \cdot 100\text{,}\) and \(100 \equiv 1 \pmod 9\) (since \(9 \mid 99\)). So we can in fact replace the 400 with simply a 4. Again, we are appealing to our claim that we can replace congruent elements, but we are really appealing to property 3 about the arithmetic of congruence: we know \(100 \equiv 1 \pmod{9}\text{,}\) so if we multiply both sides by \(4\text{,}\) we get \(400 \equiv 4 \pmod 9\text{.}\)

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.