<s>[INST] 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?
 [/INST] \begin{align}
   n \equiv 5 &\pmod{6} \\
   n \equiv 2 &\pmod{11} \\
   n \equiv 31 &\pmod{35}
\end{align}
The idea is to find three integers, $A, B,$ and $C$ such that
\begin{array}{|l|l|l|}
   A \equiv 1 \pmod{6} & 
   B \equiv 0 \pmod{6} & 
   C \equiv 0 \pmod{6}
\\
   A \equiv 0 \pmod{11} & 
   B \equiv 1 \pmod{11} &
   C \equiv 0 \pmod{11}
\\
   A \equiv 0 \pmod{35} &
   B \equiv 0 \pmod{35} &
   C \equiv 1 \pmod{35}
\end{array}
Your answer will then be $5A + 2B + 31C \pmod{2310}$ where
$2310 = 6 \cdot 11 \cdot 35$. Since $6, 11,$ and $35$ are pairwise prime, the Chinese remainder theorem guarantees that $A, B,$ and $C$ exist.
There is a formula that describes how to find $A, B,$ and $C$, but it isn't that much trouble to figure them out from scratch.
The equivalences $A \equiv 0 \pmod{11}$ and $A \equiv 0 \pmod{35}$ imply that $A$ must be a multiple of $385 = 11 \cdot 35.$ So $A = 385x$ for some integer $x$. We solve
\begin{align}
   A    &\equiv 1 \pmod 6 \\
   385x &\equiv 1 \pmod 6 \\
   x    &\equiv 1 \pmod 6 \\
\hline
   A &= 385(1) = 385
\end{align}
The equivalences $B \equiv 0 \pmod{6}$ and $B \equiv 0 \pmod{35}$ imply that $B$ must be a multiple of $210 = 6 \cdot 35.$ So $B = 210x$ for some integer $x$. We solve
\begin{align}
   B    &\equiv 1 \pmod{11} \\
   210x &\equiv 1 \pmod{11} \\
   x    &\equiv 1 \pmod{11} \\
\hline
   B &= 210(1) = 210
\end{align}
The equivalences $C \equiv 0 \pmod{6}$ and $C \equiv 0 \pmod{11}$ imply that $C$ must be a multiple of $66 = 6 \cdot 11.$  So $C = 66x$ for some integer $x$. We solve
\begin{align}
   C    &\equiv  1 \pmod{35} \\
   66x  &\equiv  1 \pmod{35} \\
   -4x  &\equiv  1 \pmod{35} \\
   x    &\equiv -9 \pmod{35} \\
\hline
   C &= 66(-9) = -594
\end{align}
We then compute
\begin{align}
   n 
      & \equiv 5A + 2B + 35C \pmod{2310} \\
      &\equiv 5(385) + 2(210) + 31(-594) \pmod{2310} \\
      &\equiv 101\pmod{2310} \\
\end{align}
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