1. Solve the following system of linear congruences:
(i) $x \equiv 1(\bmod 4), x \equiv 2(\bmod 3), x \equiv 3(\bmod 5)$;
(ii) $x \equiv 4(\bmod 11), x \equiv 3(\bmod 17)$;
(iii) $x \equiv 2(\bmod 5), x \equiv 1(\bmod 6), x \equiv 3(\bmod 7), x \equiv 0(\bmod 11)$;
(iv) $3 x \equiv 1(\bmod 11), 5 x \equiv 7(\bmod 13)$;
(v) $8 x \equiv 6(\bmod 10), 3 x \equiv 10(\bmod 17)$;
(vi) $x \equiv 7(\bmod 10), x \equiv 3(\bmod 12), x \equiv 12(\bmod 15)$;
(vii) $x \equiv 6(\bmod 35), x \equiv 11(\bmod 55), x \equiv 2(\bmod 33)$. (Solve (i), (ii), (iv), (v) using the method in Proof 2 of Theorem 1)
1. (i) $x=1+4 y, 4 y \equiv 1(\bmod 3), 4 y \equiv 2(\bmod 5)$. Thus, $y \equiv 1(\bmod 3), y \equiv -2(\bmod 5) . y=1+3 z, 3 z \equiv -3(\bmod 5), z \equiv -1(\bmod 5) . y \equiv -2(\bmod 15), x \equiv -7(\bmod 60)$.
(ii) $x=4+11 y, 11 y \equiv -1(\bmod 17), y \equiv 3(\bmod 17), x \equiv 37(\bmod 187)$.
(iii) $m_{1}=5, m_{2}=6, m_{3}=7, m_{4}=11 . M_{1}=6 \cdot 7 \cdot 11 \equiv 2(\bmod 5), 3 M_{1} \equiv 1\left(\bmod m_{1}\right), M_{2}=5 \cdot 7 \cdot 11 \equiv 1(\bmod 6), 1 \cdot M_{2} \equiv 1\left(\bmod m_{2}\right) . M_{3}=5 \cdot 6 \cdot 11 \equiv 1(\bmod 7), 1 \cdot M_{3} \equiv 1\left(\bmod m_{3}\right) . M_{4}=5 \cdot 6 \cdot 7 \equiv 1(\bmod 11), 1 \cdot M_{4} \equiv 1(\bmod 11)$. Therefore, $x \equiv 3 \cdot 6 \cdot 7 \cdot 11 \cdot 2 + 5 \cdot 7 \cdot 11 \cdot 1 + 5 \cdot 6 \cdot 11 \cdot 3 (\bmod 5 \cdot 6 \cdot 7 \cdot 11)$.
(iv) $x=4+11 y, 55 y \equiv 0(\bmod 13), x \equiv 4(\bmod 143)$.
(v) $x=-3+5 y, 15 y \equiv 2(\bmod 17), y \equiv -1(\bmod 17), x \equiv -8(\bmod 85), x \equiv -8, 77(\bmod 170)$.
(vi) $x \equiv 27(\bmod 60)$. (vii) No solution. The second and third equations are contradictory.