Chinese remainder theorem #2018

• $5 \mid 2x - 2 = 2(x - 1) \iff 5 \mid x - 1$ because $\text{gcd(2,5) = 1}$

• $15 \mid 10x - 5 \iff 5 \cdot 3 \mid 5(2x - 1) \iff 3 \mid 2x - 1 \iff 3 \mid 2x - 4 = 2(x - 2) \iff 3 \mid x - 2$ because $\text{gcd(2,3) =1}$ .

Because of this $A$ has the same solutions than $\displaystyle B \begin{cases} x \equiv 1 \space \text{mod 2} \\ x \equiv 1 \space \text{mod 5} \\ x \equiv 2 \space \text{mod 3} \end{cases} \iff C \begin{cases} x \equiv 1 \space \text{mod 10} \\ x \equiv 2 \space \text{mod 3} \end{cases}$ Applying chinese remainder theorem one solution of $C$ is $x = 11 = 1 \cdot 3 \cdot (-3) + 2 \cdot 10 \cdot 1$ and therefore the general solutions of $A$ are $x = 11 + 30 \cdot \lambda$ with $\lambda \in \mathbb{Z}$ so $a = 11$ and $b = 11 + 30 \cdot 32 = 971$ and then $a + b = 982$