x≡1 (mod 8) x≡9 (mod 12) has solution x = x_0. How many solutions mod 24 are there to the system of congruences?

Question

Say,

x≡1 (mod 8)

x≡9 (mod 12)

has solution x = x_0. How many solutions mod 24 are there to the system of congruences?

I am worried this question is too easy to be true. That is why I am confused. Don't know where to start.

Answer 1

$$x\equiv1\pmod8\equiv9$$ and $$x\equiv9\pmod{12}$$

$$\implies x\equiv9\pmod{\text{lcm}(8,12)}$$

Answer 2

Because $gcd(8, 12) = 4 \neq 1$, the Chinese remainder theorem doesn't work.
There may or may not be a solution.

First we can check, if there is a solution. I would split the second congruence into 2 congruences. $$ x \equiv 1 \text{ mod } 4\\x \equiv 0 \text{ mod } 3$$

Because $ x \equiv 1 \text{ mod } 8$ also implies $ x \equiv 1 \text{ mod } 4$, the set of congruences is solvable. And the chinese remainder theorem tells us, that there is one solution $\text{mod } gcd(8,3)$.

Answer 3

If $\,\color{#c00}{a\equiv b}\pmod n\,$ then $\ \begin{eqnarray}x\equiv \color{#c00}a\!\!\!\pmod n\\ x\equiv b\!\!\!\pmod{\! m}\end{eqnarray}$ $\!\iff\!$ $\begin{eqnarray}x\equiv \color{#c00}b\!\!\!\pmod n\\ x\equiv b\!\!\!\pmod{\! m}\end{eqnarray}$ $\!\color{#0a0}\iff\!$ $x\equiv b\pmod{\!{\rm lcm}(n,m)}$

where, above, we have employed the basic lcm law: $\,\ n,m\mid x-b \color{#0a0}\iff {\rm lcm}(n,m)\mid x-b$

Remark $\ $ This is a special case of the uniquitous constant case optimization of CRT. You can find many examples of this and further discussion in various prior posts.