Solve the system of linear congruences $x \equiv 2 ~(mod ~3)$ , $x \equiv 3 ~(mod ~5)$, $x \equiv 2 ~(mod ~7)$.

Example: Solve the system of linear congruences

$x \equiv 2 ~(mod ~3)$,

$x \equiv 3 ~(mod ~5)$,

$x \equiv 2 ~(mod ~7)$.

Answer: Let $ n = 3 \cdot 5 \cdot 7 = 105, ~N_{1} = \dfrac{n}{3} = \dfrac{105}{3} = 35, ~ N_2 = \dfrac{n}{5} = \dfrac{105}{5} = 21, ~ N_3 = \dfrac{n}{7} = \dfrac{105}{7} = 15$.

Now, we find solution of linear congruences

$35 x_{1} \equiv 1 ~(mod ~3), ~ 2 1 x_2 \equiv 1 ~(mod ~5),  15x_3 \equiv 1 ~(mod ~7)$

We will find solution of linear congruences $35 x_{1} \equiv 1 ~(mod ~3), ~ 2 1 x_2 \equiv 1 ~(mod ~5),  15x_3 \equiv 1 ~(mod ~7)$ by trial and error method.  Firstly we find solution of linear congruence $35 x_{1} \equiv 1 ~(mod ~3)$. We will consider simple values of $x_{1}$ as $1, ~-1,~ 2,~ -2,~ \cdots $ and check which value satisfy given linear congruence $35 x_{1} \equiv 1 ~(mod ~3)$. When we take $x_{1} = 1, ~-1,~ 2,~ -2,~3, ~ -3$, we observe that $x_{1} =1, ~-1,$ are not a solution of $35 x_{1} \equiv 1 ~(mod ~3)$. Now, we take $x_{1} = 2$, we observe that $x_{1} = 2$ is  a solution of $35 x_{1} \equiv 1 ~(mod ~3)$. 

We use same method to find solution of $ 21 x_2 \equiv 1 ~(mod ~5),  15x_3 \equiv 1 ~(mod ~7)$. We observed that $x_2 = 1$ is a solution of $ 21x_2 \equiv 1 ~(mod ~5)$, also $x_3 = 1$ is a solution of  $  15x_3 \equiv 1 ~(mod ~7)$. Thus, a solution of the system of given linear congruences is given by 

$x = 2 \cdot 35 \cdot 2 + 3 \cdot 21 \cdot 1 + 2 \cdot 15 \cdot 1 = 233 $

modulo $105$, we get the unique solution $x = 233 \equiv 23 ~(mod ~105)$.

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