Solution of a system of congruence equations Consider the following system out of which I want to find all possible values of $\lambda$.
\begin{eqnarray}
1287\lambda\equiv0 (\mathrm{mod}\ 6)\label{eq1}\\
165\lambda\equiv0 (\mathrm{mod}\ 4)\label{eq2}\\
9\lambda\equiv0 (\mathrm{mod}\ 2)\label{eq3}
\end{eqnarray}
Since I am not comfortable with the Chinese Remainder Theorem, I have tried to proceed with direct substitution. 
\begin{eqnarray}
1287\lambda=6k\Rightarrow\lambda=\frac{2k}{429}\label{eq4}\\
165\frac{2k}{429}=4r\Rightarrow k=\frac{858r}{165}\label{eq5}\\
\lambda=\frac{4r}{165}\label{eq6}\\
9\lambda=2t\Rightarrow r=\frac{55t}{6}\label{eq7}\\
\lambda=\frac{2t}{9}\\
\end{eqnarray}
i.e. we seek $t$ such that $2t\equiv0 (\mathrm{mod}\ 9)\Rightarrow t\equiv0 (\mathrm{mod}\ 9)$ i.e. $t=9,18,\dots$. Thus, the possible values of $\lambda$ are $\lambda=2,4,\dots$
I have definitely done something wrong here since e.g. $\lambda=2$ cannot work, but I cannot spot the mistake. Alternative ways are also welcome.

$\!\bmod 6\!:\ \ 0 \equiv 1287 \lambda \equiv 3 \lambda\iff 6\mid 3\lambda\iff 2\mid \lambda$
$\!\bmod 4\!:\ \ 0\equiv 165 \lambda\equiv \lambda\iff 4\mid \lambda$
$\!\bmod 2\!:\ \ 0\ \equiv\ 9\lambda\ \equiv\ \lambda\iff 2\mid \lambda$
So the system is equivalent to $\,2,4\mid \lambda\iff 4\mid \lambda$
