Smallest positive integer that gives remainder 5 when divided by 6, remainder 2 when divided by 11, and remainder 31 when divided by 35? What is the smallest positive integer that gives the remainder 5 when divided by 6, gives the remainder 2 when divided by 11 and 31 when divided by 35? 
Also, are there any standard methods to solve problems like these?

The system of equations
$$ \left\{\begin{array}{ll} n \equiv 5 &\pmod{6} \\ n \equiv 2&\pmod{11} \\ n\equiv 31 &\pmod{35}\end{array}\right. $$
is equivalent to $n\equiv m\pmod{2\cdot 3\cdot 5\cdot 7\cdot 11}$ by the Chinese remainder theorem.
If $m_6, m_{11}, m_{35}$ are the smallest integer solutions of the systems
$$ \left\{\begin{array}{ll} n \equiv 1 &\pmod{6} \\ n \equiv 0&\pmod{11} \\ n\equiv 0 &\pmod{35}\end{array}\right. \quad\left\{\begin{array}{ll} n \equiv 0 &\pmod{6} \\ n \equiv 1&\pmod{11} \\ n\equiv 0 &\pmod{35}\end{array}\right.\quad \left\{\begin{array}{ll} n \equiv 0 &\pmod{6} \\ n \equiv 0&\pmod{11} \\ n\equiv 1 &\pmod{35}\end{array}\right.$$
then we have $m\equiv 5m_6+2m_{11}+31 m_{35}\pmod{2310}$, and $m_6,m_{11},m_{35}$ are not difficult to compute. For instance, $m_6$ is a multiple of $11\cdot 35$ that is $\equiv 1\pmod{6}$, but
$$ k\cdot 11\cdot 35 \equiv 1\pmod{6} $$
is equivalent to $k\equiv 1\pmod{6}$, hence $m_6=11\cdot 35=385$. In a similar way you may find $m_{11}=210$ and $m_{35}=1716$, then re-combine them in order to find $m=\color{red}{101}$.
