What does the minimum x that satify $x=24n+12=15m+6=11k+2$? My attempt so far was: let $x=24n+12=15m+6=11k+2$ find $x$ as the form $x=24*15*a+24*11*b+15*11*c$ $$2\equiv 24\cdot 15\cdot a\quad (mod\quad 11)\quad \Rightarrow 1\equiv 4\cdot a\Rightarrow \quad a=3\\\ 6\equiv 24\cdot 11\cdot b\quad (mod\quad 15)\quad \Rightarrow \quad 2\equiv 8\cdot 11\cdot b\quad (mod\quad 5)\quad \Rightarrow 2\equiv 3\cdot b\quad (mod\quad 5)\quad \quad $$ and I stack here after finding $a$

$\\!x\equiv 12\pmod{\\!24}\\!\iff x/3\,\equiv\ \ \ 4\,\pmod{\\! 8}$
$\\!\\!\left.\begin{align} &x\equiv \ 6\\!\\!\pmod{\\!15}\\!\iff x/3\,\equiv\ \ \ 2\\!\\!\\!\pmod{ 5}\\\ &x\equiv \ 2\\!\\!\pmod{\\!11}\\!\iff x/3\,\equiv -3\\!\\!\\!\pmod{\\!\\!11}\\\ \end{align}\right\\}\\!\\!\\!\iff\\!\dfrac{x}3\equiv -3\pmod{\\!55}\\!\iff\\! \dfrac{x}3 = \color{#0a0}{-3\\!+\\!55j}$

$\\!\\!\\!\bmod \color{#c00}8\\!:\, 4\equiv \dfrac{x}3\equiv \color{#0a0}{-3\\!+\\!55}\color{#c00}j\equiv 5\\!-\\!j\\!\\!\iff\\! \color{#c00}{j\equiv 1}\\!\iff\\! \dfrac{x}3=-3\\!+\\!55(\color{#c00}{1\\!+\\!8i})\\!\iff\\!\\! x\equiv 156\\!+\\!1320i $