Modular Arithmetic - Pirate Problem

proof-writing modular-arithmetic

The innocent way. Reduce your work with the $\mathrm{\color{Red}{red}}$ equivalence between congruence systems.

$$\begin{cases} x\equiv 2\pmod 7 \\ x\equiv 5 \pmod 6 \\ x \equiv 2 \pmod 5 \end{cases}\color{Red}{\iff}\begin{cases} x\equiv 2\pmod {35} \\ x\equiv 5 \pmod 6 \end{cases}$$

From the first congruence, $x=35t+2$ for some $t$. This implies $$35t+2\equiv 2-t\equiv5\pmod 6\iff t\equiv 3\pmod 6.$$ Therefore, $t=6u+3$ for some $u$. All the solutions are then of the form $$x=35t+2=35(6u+3)+2=210u+107.$$ The least positive solution is $x_0=107.$

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