Use the Chinese remainder theorem to find the general solution of $x \equiv a \pmod {2^3}, \; x \equiv b \pmod {3^2}, \; x \equiv c \pmod {11}$ Help! Midterm exam is coming, but i still unable to solve this simple problem using the Chinese remainder theorem.
$$x \equiv a \pmod {2^3}, \quad x \equiv b \pmod {3^2}, \quad x \equiv c \pmod {11}.$$

We know that $x$ has the form:
$$
x=a\cdot(s\cdot 3^2\cdot 11)+b\cdot(t\cdot 2^3\cdot 11)+c\cdot(r\cdot 2^3\cdot 3^2)
$$
where $s,t$ and $r$ have been found so that
$$
\begin{align}
s\cdot 3^2\cdot 11&\equiv 1\mbox{ mod }2^3\\
t\cdot 2^3\cdot 11&\equiv 1\mbox{ mod }3^2\\
r\cdot 2^3\cdot 3^2&\equiv 1\mbox{ mod }11
\end{align}
$$
Just try $s=1$, $s=2$ etc. until you find $s$ and similarly for $t$ and $r$. Or use Bezout's identity from the extended Eucledian algorithm to find $s,t$ and $r$.
