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

From the condition of the equation we have:
$$x \equiv a \pmod 8 \implies x = 8k + a$$
$$x \equiv b \pmod 9 \implies x = 9n + b$$
$$x \equiv c \pmod {11} \implies x = 11m + c$$
Now we have:
$$8k+a=9n+b$$
$$8k+a\equiv b \pmod 9$$
$$8k\equiv b-a \pmod 9$$
Having actual values would be easier to get congruence relation for k modulo 9, but now we'll use:
$$k \equiv \frac{b-a + 9s}{8} \pmod 9 \implies k = 9t + \frac{b-a+9s}{8}$$
Note that if we add 9s on the RHS the congruence relation won't change. Sowe add $9s$ in order to get an integer when we divide by 9.
Now substitute back we have:
$$x=8k+a = 8\left(9t + \frac{b-a}{8}\right) = 72t + b-a + 9s$$
Now using this relation for x we do the same thing for this realtion and $x = 11m + c$. It maybe clearer to you with example:

$$x \equiv 3 \pmod 8 \implies x = 8k + 3$$
$$x \equiv 2 \pmod 9 \implies x = 9n + 2$$
$$x \equiv 5 \pmod {11} \implies x = 11m + 5$$
$$8k + 3 = 9n + 2$$
$$8k + 3 \equiv 2 \pmod 9$$
$$8k \equiv -1 \equiv 8 \pmod 9$$
$$k \equiv 1 \pmod 9 \implies k = 9t + 1$$
Now we substitute back:
$$x = 8k + 3 = 8(9t+1) + 3 = 72t + 8 + 3 = 72t + 11$$
Now we repeat the same procedure:
$$72t + 11 = 11m + 5$$
$$72t + 11 \equiv 5 \pmod {11}$$
$$72t \equiv 5  \equiv 720 \pmod {11}$$
$$t \equiv 10 \pmod {11} \implies t = 11s + 10$$
Now we substitue:
$$x = 72t + 11 = 72(11s + 10) + 11 = 792s + 720 + 11 = 792s + 731$$
We have a congruence relation for $x$:
$$x \equiv 731 \pmod{792}$$
You can see that $792=8\cdot 9 \cdot 11$, that's because all moduli are coprime, otherwise we would end up with the least common multiple of the moduli.
