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We choose 17 because \(17x\) will have remainder 0. This will allow us to reduce the congruence to just one variable. We could have also moved to a congruence modulo 29, although there is usually a good reason to select the smaller choice, as this will allow us to reduce the other coefficient. In our case, we reduce the congruence as follows:

\begin{equation*} \begin{aligned}17x + 29y \amp \equiv 41 \pmod{17} \\ 0x + 12y \amp \equiv 7 \pmod{17} \\ 12 y \amp \equiv 24 \pmod{17} \\ y \amp \equiv 2 \pmod{17}. \end{aligned} \end{equation*}

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