同余式组求解

中国剩余定理

\left\{ \begin{array}{**lr**} x\equiv b_1\left(\mod m_1\right) &\\ \vdots &\\ x\equiv b_k\left(\mod m_k\right) \end{array} \right.

令$m=m_1\cdot m_2\cdots m_k=m_i\cdot M_i$ $M_{i}^{\prime}\cdot M_i\equiv1\left(\mod m_i\right), i=1,\cdots,k$ $x\equiv \sum\limits_{i=1}^{k}{b_i\cdot M_{i}^{\prime}\cdot M_i}\left(\mod m\right)$