Solve the following simultaneous equations:
(a) $x ≡ 5 \pmod 7 $, $x ≡ 4 \pmod {11}$
(b) $x ≡ 0 \pmod 5$, $x ≡ 0 \pmod 6$
(c) $x ≡ 3 \pmod 8$, $x ≡ 5 \pmod {13}$
(d) $x ≡ 1 \pmod 3$, $x ≡ 2 \pmod 5$, $x ≡ 3 \pmod 7$
(e) $x ≡ 1 \pmod 5$, $x ≡ 3 \pmod 7$, $x ≡ 5 \pmod {11}$
(a) The modulii 7 and 11 are indeed pair-wise coprime, so we can use the Chinese Remainder Theorem to solve
$x ≡ 5 \pmod 7 $
$x ≡ 4 \pmod {11}$
Formula (3.23) here is
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 = 5(11)x_1 + 4(7)x_2$$
where $N_k x_k \equiv 1 \pmod {n_k}$, and so
$$ \begin{align} 11 x_1 & \equiv 1 \pmod 7 \\ \\ 4 x_1 & \equiv 1 \pmod 7 \\ \\ x_1 & \equiv 2 \pmod 7 \end{align}$$
and
$$ \begin{align} 7 x_2 & \equiv 1 \pmod {11} \\ \\ x_2 & \equiv 8 \pmod {11} \end{align}$$
which gives
$$ x = a_1N_1x_1 + a_2N_2x_2 = 5(11)2 + 4(7)8 = 334 $$
The general unique solution is
$$ \begin{align} x & \equiv 334 \pmod {7 \times 11} \\ \\ x & \equiv 26 \pmod {77} \end{align}$$
(b) The modulii 5 and 6 are indeed pair-wise coprime., so we can use the Chinese Remainder Theorem to solve
$x ≡ 0 \pmod 5$
$x ≡ 0 \pmod 6$
Formula (3.23) here is
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 = 0 \quad \text { since } a_1=0, a_2 = 0$$
So the general unique solution is
$$ \begin{align} x & \equiv 0 \pmod {5 \times 6} \\ \\ x & \equiv 0 \pmod {30} \end{align}$$
(c) The modulii 8 and 13 are indeed pair-wise coprime, so we can use the Chinese Remainder Theorem to solve
$x ≡ 3 \pmod 8$
$x ≡ 5 \pmod {13}$
Formula (3.23) here is
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 = 3(13)x_1 + 5(8)x_2$$
where $N_k x_k \equiv 1 \pmod {n_k}$, and so
$$ \begin{align} 13 x_1 & \equiv 1 \pmod {8} \\ \\ 5 x_1 & \equiv 1 \pmod 8 \\ \\ x_1 & \equiv 5 \pmod 8 \end{align}$$
and
$$ \begin{align} 8 x_2 & \equiv 1 \pmod {13} \\ \\ x_2 & \equiv 5 \pmod {13} \end{align}$$
which gives
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 = 3(13)(5) + 5(8)(5) = 395 $$
So the general unique solution is
$$ \begin{align} x & \equiv 395 \pmod {8 \times 13} \\ \\ x & \equiv 83 \pmod {104} \end{align}$$
(d) The modulii 3, 5 and 7 are indeed pair-wise coprime, so we can use the Chinese Remainder Theorem to solve
$x ≡ 1 \pmod 3$
$x ≡ 2 \pmod 5$
$x ≡ 3 \pmod 7$
Formula (3.23) here is
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 = 1(5 \times 7)x_1 + 2(3 \times 7)x_2 + 3(3 \times 5)x_3 $$
where $N_k x_k \equiv 1 \pmod {n_k}$, and so
$$ \begin{align} 35 x_1 & \equiv 1 \pmod 3 \\ \\ 2 x_1 & \equiv 1 \pmod 3 \\ \\ x_1 & \equiv 2 \pmod 3 \end{align}$$
and
$$ \begin{align} 21 x_1 & \equiv 1 \pmod 5 \\ \\ x_1 & \equiv 1 \pmod 5 \end{align}$$
also
$$ \begin{align} 15 x_3 & \equiv 1 \pmod 7 \\ \\ x_3 & \equiv 1 \pmod 7 \end{align}$$
which gives
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 = 1(5 \times 7)2 + 2(3 \times 7)1 + 3(3 \times 5)1 = 157 $$
So the general unique solution is
$$ \begin{align} x & \equiv 157 \pmod {3 \times 5 \times 7} \\ \\ x & \equiv 52 \pmod {105} \end{align}$$
(e) The modulii 5, 7 and 11 are indeed pair-wise coprime, so we can use the Chinese Remainder Theorem to solve
$x ≡ 1 \pmod 5$
$x ≡ 3 \pmod 7$
$x ≡ 5 \pmod {11}$
Formula (3.23) here is
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 = 1(7 \times 11)x_1 + 3(5 \times 11)x_2 + 5(5 \times 7)x_3 $$
where $N_k x_k \equiv 1 \pmod {n_k}$, and so
$$ \begin{align} 77 x_1 & \equiv 1 \pmod 5 \\ \\ 2 x_1 & \equiv 1 \pmod 5 \\ \\ x_1 & \equiv 3 \pmod 5 \end{align}$$
and
$$ \begin{align} 55 x_2 & \equiv 1 \pmod 7 \\ \\ 6 x_2 & \equiv 1 \pmod 7 \\ \\ x_2 & \equiv 6 \pmod 7 \end{align}$$
also
$$ \begin{align} 35 x_3 & \equiv 1 \pmod {11} \\ \\ 2 x_3 & \equiv 1 \pmod {11} \\ \\ x_3 & \equiv 6 \pmod {11} \end{align}$$
which gives
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 = 1(7 \times 11)3 + 3(5 \times 11)6 + 5(5 \times 7)6 = 2271 $$
So the general unique solution is
$$ \begin{align} x & \equiv 2271 \pmod {5 \times 7 \times 11} \\ \\ x & \equiv 346 \pmod {385} \end{align}$$