$\begin{array}{l}\text{- Ta có : $a \div 12$ dư $8 \to a=12k+8^{(*)}\,\,(k\in\mathbb{N})$}\\\text{- Ta lại có : $a \div 19$ dư $10 \to a-10\,\,\vdots\,\,19$}\\\to12k+8-10\,\,\vdots\,\,19\\\to12k-2\,\,\vdots\,\,19\\\to12k-2+38\,\,\vdots\,\,19\\\to12k+36\,\,\vdots\,\,19\\\to 12(k+3)\,\,\vdots\,\,19\\\text{mà $(12;19)=1$}\\\to k+3\,\,\vdots\,\,19\\\to k+3=19m\,\,(m \in\mathbb{N})\\\to k=19m-3\\\text{- Thay $k=19m-3$ vào $(*)$ ta có :}\\a=12k+8\\\to a=12(19m-3)+8\\\to a=228m-36+8\\\to a=228m-28\\\to a-200=228m-28-200\\\to a-200=228m-228\,\,\vdots\,\,228\\\to a-200\,\,\vdots\,\,228\\\to a-200=228n\,\,(n\in\mathbb{N})\\\to a=228n+200\\\to \text{$a \div 228$ dư 200}\\\text{- Vậy $a \div 228$ dư 200} \end{array}$
