Find the least positive integer $x$ such that $x \equiv 5 ~(mod ~7)$ , $x \equiv 7 ~(mod ~11)$ , $x \equiv 3 ~(mod ~13)$.
Example: Find the least positive integer $x$ such that $x \equiv 5 ~(mod ~7)$, $x \equiv 7 ~(mod ~11)$, $x \equiv 3 ~(mod ~13)$. Answer: Let $ n = 7 \cdot 11 \cdot 13 = 1001, ~N_{1} = \dfrac{n}{7} = \dfrac{1001}{7} = 143, ~ N_2 = \dfrac{n}{11} = \dfrac{1001}{11} = 91, ~ N_3 = \dfrac{n}{13} = \dfrac{1001}{13} = 77$. Now, we find solution of linear congruences $143 x_{1} \equiv 1 ~(mod ~7), ~ 91 x_2 \equiv 1 ~(mod ~11), 77x_3 \equiv 1 ~(mod ~13)$ We will find solution of linear congruences $143x_{1} \equiv 1 ~(mod ~7), ~ 91 x_2 \equiv 1 ~(mod ~11), 77x_3 \equiv 1 ~(mod ~13)$ by trial and error method. Firstly we find solution of linear congruence $143 x_{1} \equiv 1 ~(mod ~7)$. We will consider simple values of $x_{1}$ as $1, ~-1,~ 2,~ -2,~ \cdots $ and check which value satisfy given linear congruence $143 x_{1} \equiv 1 ~(mod ~7)$. When we take $x_{1} = 1, ~-1,~ 2,~ -2,~3, ~ -3$, we observe that $x_{1} =1, ~-1,~ ...