Solve the linear congruence $18x \equiv 30 ~(mod ~42)$.

Example: Solve the linear congruence $18x \equiv 30 ~(mod ~42)$.

 Answer: Since $gcd(18, 42) = 6$ and $6$ divides $30$, by  Theorem, given linear congruence $18x \equiv 30 ~(mod ~42)$ has $6$ mutually incongruent solutions modulo $42$.  We will find first solution by trial and error method. We will consider simple values of $x$ as $1, ~-1,~ 2,~ -2,~ \cdots $ and check which value satisfy given linear congruence $18x \equiv 30 ~(mod ~42)$. When we take $x = 1, ~-1,~ 2,~ -2,~3, ~ -3$, we observe that $x =1, ~-1,~ 2,~ -2,~3$ are not a solution of $18x \equiv 30 ~(mod ~42)$. Now, we take $x = -3$, we observe that $x = -3$ is  a solution of $18x \equiv 30 ~(mod ~42)$. But $-3 \equiv 39 ~ (mod~ 42)$, So $x_{0} = 39$ is a solution of a given  linear congruence $18x \equiv 30 ~(mod ~42)$.  By Theorem, other solutions of given linear congruence $18x \equiv 30 ~(mod ~42)$  are given by 

$x_{1} = 39 + \frac{42}{6} = 39 + 7 = 46 \equiv 4~ (mod ~ 42)$, 

 $x_2 = 39 + (\frac{42}{6}) ~2 = 39 + 7 \cdot 2 = 39 +14 = 53 \equiv 11 ~(mod ~ 42)$,

 $x_3 = 39 + (\frac{42}{6}) ~3 = 39 + 7 \cdot 3 = 39 + 21 = 60 \equiv 18 ~(mod ~ 42)$, 

$x_4 = 39 + (\frac{42}{6}) ~4 = 39 + 7 \cdot 4 = 67 \equiv 25 ~(mod ~ 42)$, 

$ x_5 = 39 + (\frac{42}{6}) ~5 = 39 + 7 \cdot 5 = 39 + 35 = 74 \equiv 32 ~(mod ~ 42)$. 

Therefore $4, ~ 11, ~18, ~25, ~32$ and $ 39$ are $6$ mutually incongruent solutions modulo $42$ of  given linear congruence $18x \equiv 30 ~(mod ~42)$.


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