Congruence with powers

Question

How would one go about solving the following congruence:

$x\equiv2^{2012} \pmod6$

I assume you can divide that number on the left to lower it, but how exactly?

Same issue applies to $x\equiv6^{29} \pmod7$

Answer 1

Hints:

$2^3\equiv2\mod6$

$6^2\equiv1\mod7$

Answer 2

I'd like to suggest you a general way to approach $x \equiv 2^{2012} (mod 6)$.

Let's notice that a congruence $mod 6$ it implies a congruence $mod3,mod2$ by Chinese remainder theorem,which also guaranties unique solution of the congruence $mod 6$.

Studying $x\equiv 2^{2012} (mod2)$ it gives us $x \equiv 0 \hspace{0.1cm} (mod2),$ or, in other words $x$ even.

Hence exists$ y \in \mathbb{Z}:x=2y$,

Now we have : $$2y\equiv 2^{2012} \hspace{0.1cm} (mod6)$$

Just now we can divide by 2, obtaining $$y \equiv 2^{2011} \hspace{0.1cm} (mod3)$$

By Fermat's little theorem we know that $2^{3-1} \equiv 1 (mod3)$; Noticing that 2011 = 2 $\cdot 1005+1$ we get $2^{2011} = 2^{2 \cdot 1005} \cdot 2 \equiv 2 \hspace{0.1cm} (mod3)$

We almost done, remembering that $x=2y$ we obtain :

$$y \equiv 2 \hspace{0.1cm} (mod3) \rightarrow x = 2\cdot y \equiv 2 \cdot 2 = 4 \hspace{0.1cm} mod(6)$$

Solution of the problem.

Let me know if there's anything wrong or something that doesn't feel right to you.