- Find the smallest positive solution \(x\) of the system of congruences
\[\begin{aligned}
x&\equiv 4(\mod 3),\\
x&\equiv5(\mod 7),\\
x&\equiv6(\mod 11).
\end{aligned}
\]
Solution: Using the Chinese Remainder Theorem. \(m=3\times7\times11=231\Rightarrow \hat m_1=\frac{231}{3}=77,\hat m_1=\frac{231}{7}=33,\hat m_3=\frac{231}{11}=21\).
- \(77\equiv 2(\mod 3),~2\times 2\equiv1(\mod 3)\Rightarrow l_1=2\);
- \(33\equiv 5(\mod 7),~5\times3\equiv1(\mod7)\Rightarrow l_2=3\);
- \(21\equiv10(\mod 11),~10\times10\equiv1(\mod 11)\Rightarrow l_3=10\).
Put \(x=4\times77\times2+5\times33\times3+6\times21\times10(\mod231)=2371(\mod 231)=61(\mod 231)\). Thus the smallest positive solution \(x=61\). #