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.
$$x \equiv a \pmod {2^3}, \quad x \equiv b \pmod {3^2}, \quad x \equiv c \pmod {11}.$$

The general method to find such a number, is to solve for $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$, where the numbers in brackets represent a number modulo $8$, $9$, $11$.  Since the number $(1,1,1) = 1$, it is necessary to solve only for two of them.
For $(1,0,0)$, we start off with $99 = (3, 0, 0)$.  Since the first multiple of $3$ to exceed eight by 1, is 9, we then have  $3\times 99 = (1,0,0)$.
For $(0,1,0)$, we start with $8\times 11 = (0,7,0)$.  It then is a matter of multiplying this by $4$ to get $352 = (0,1,0)$
We could do the same for the next, but note that $(1,1,1)=1+792$, and thence $(0,0,1) = (1,1,1)-(1,0,0)-(0,1,0)$, gives $793 - 297 - 352 = 144$  This is indeed $(0,0,1)$.
So we find here that $a=297$, $b=352$ and $c=144$, the sum is $1 \pmod{8\cdot 9 \cdot 11}$.
