Congruence with power

Question

We were given a congruence $$ X=2^{2012} \mod{6} $$ And we asked our professor what's the solution and she replied with $$ X=2^{2012} \mod{2} $$ $$ X=0^{2012} \mod{2} $$ $$ X=0 \mod{2} $$ My question is how did she get from mod6 to mod2, is there an actual rule for this or is it a mistake?

Answer 1

By CRT: $\ \ x\equiv 2^{\large 2N}\!\pmod{\!6}\iff \begin{align} &x\equiv 2^{\large 2N}\!\equiv 0\!\! \pmod{\!2}\ \ \ {\rm by}\ \ N > 0\ \ ({\rm here}\ 2N = 2012)\\ &x\equiv \color{#c00}2^{\large\color{#c00} 2N}\!\equiv 1\!\!\pmod{\!3}\ \ \ {\rm by}\ \ (\color{#c00}{2^{\large 2}})^N\equiv \color{#c00}1^N\!\equiv 1 \end{align}\ \ $

But $\,x\equiv 1\pmod{\!3}$ is redundant (same as first congruence in system) leaving only $\,x\equiv 0\pmod{\!2}$