What does the minimum x that satify $x=24n+12=15m+6=11k+2$?

Question

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 $a$

Answer 1

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

$\!\!\!\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 $

Answer 2

\begin{align} 24n + 12 &= 15m + 6 \\ 8n + 4 &= 5m + 2 \\ 5m - 8n &= 2 \\ \end{align}

Since $10-8=2$ we see that $(m,n)=(2,1)$ is a particular solution. So, in general $(m,n) = (2+8t, 1+5t)$.

Then \begin{align} 15m + 6 &= 11k + 2 \\ 15(2+8t) + 6 &= 11k + 2 \\ 120t + 36 &= 11k + 2 \\ 11k - 120t &= 34 \\ \end{align}

Noting that $120+34 = 154 = 14 \times 11$, we see that $(k,t) = (14,2)$ is a particular solution. So, in general, $(k, t) = (14+120u, 2+11u)$.

So, in general, $$\begin{array}{l} m = 2+8t = 88u + 18 \\ n = 1+5t = 55u + 11 \\ k = 120u + 14 \end{array}$$

We then find that $$\begin{array}{l} x = 24n + 12 = 1320 u + 276 \\ x = 15m + 6 = 1320 u + 276 \\ x = 11k + 2 = 1320 u + 156 \end{array}$$

The smallest positive value of $x$ is $1320$.

If you want to use congruences, then

\begin{align}\ x = 24n + 12 &\iff x \equiv 12 \pmod{24} &\iff \left\{\begin{array}{l} x \equiv 4 \pmod 8 \\ x \equiv 0 \pmod 3 \end{array} \right.\\ x = 15m + 6 &\iff x \equiv 6 \pmod{15} &\iff \left\{\begin{array}{l} x \equiv 1 \pmod 5 \\ x \equiv 0 \pmod 3 \end{array} \right.\\ x = 11k + 2 &&\iff x \equiv 2 \pmod{11} \end{align}

Which simplifies to

\begin{align}\ x &\equiv 4 \pmod{8} \\ x &\equiv 1 \pmod{5} \\ x &\equiv 0 \pmod{3} \\ x &\equiv 2 \pmod{11} \end{align}

We find $N = 8\times 5\times 3 \times 11 = 1320$; so $\dfrac N8 = 165, \dfrac N5 = 264, \dfrac N3 = 440$, and $\dfrac{N}{11} = 120$. We compute

\begin{array}{r|cccc|l} & \pmod 8 & \pmod 5 & \pmod 3 & \pmod{11}\\ \hline 165 & 5 & 0 & 0 & 0 \\ 264 & 0 & 4 & 0 & 0 \\ 440 & 0 & 0 & 2 & 0 \\ 120 & 0 & 0 & 0 & -1 \\ \hline 660 & 4 & 0 & 0 & 0 &5\times 4 \equiv 4 \pmod 8\\ -264 & 0 & 1 & 0 & 0 &4\times -1 \equiv 1 \pmod 5\\ 0 & 0 & 0 & 0 & 0 &2\times 0 \equiv 0 \pmod 3\\ -240 & 0 & 0 & 0 & 2 &-2\times -1 \equiv 0 \pmod{11}\\ \hline \end{array}

Adding, we get $x \equiv 660-264+0-240 \equiv 156 \pmod{1320}$