Compute $2^{287} \pmod 6$ Compute $2^{287} \pmod 6$.
So I have been using the technique of finding the pattern in modulus to make computing high powers in modulus easier. 
For example $4^{128} \pmod {13}$ pattern length is 6, or when the modulus numbers will start to repeat. Since $6 x 21 + 2 = 128$, I can take $(4^6)^{21} \cdot (4^2)$ and come out with an equivalence relation of $3 \pmod{13}$.
For this particular example however, the remainder of all powers of 2 in modulus 6 are 2. Thus the pattern length is 1 and I'm hitting a roadblock in solving this. Is there a different approach or am I thinking about this wrong? 
Thank you for your help.
 A: We can see that the powers of $2$ have this pattern: $2^1 \equiv 2$, $2^2 \equiv -2$, $2^3 \equiv 2$, and thus we start again. In conlusion odd powers are equivalent to $2$ and even powers are equivalent to $-2 \equiv 4$. We conclude $$2^{287} \equiv 2 (mod 6).$$
Otherwise you can see that $2^{287}$ must be even so it can be $0,2,4$, but it can't be $0$ cause it is not divisible by $3$. Now we can show that $2^{287}-4$ is not divisible by $3$ and conclude against $4$: $$2^{287}-4 \equiv(-1)^{287}-1\equiv-2\not = 0 (mod 3).$$
A: Hint: $2^n \equiv 0 \pmod2$ and $2^n \equiv (-1)^n \pmod3$ for all $n \in \Bbb N$

 When $n = 287$, it's odd, so $2^{287} \equiv -1 \pmod3$.  This gives $2^{287} \equiv 2 \pmod 6$

A: Note that $$2^3\equiv2 \pmod 6 \implies2^6\equiv2 \pmod 6\implies...2^{287}=2^{3\cdot95+2}\equiv2\cdot2^2\equiv2 \pmod6$$
A: With the Chinese remainder theorem:
$2^{287} \equiv 0\mod 3$, $\;2^{287}\equiv (-1)^{287}=-1\mod 3$. AAS $3-2=1$ is a Bézout's relation between $2$ and $3$, we deduce 
$$2^{287} \equiv 0\cdot3-(-1)\cdot 2=2\mod 2\cdot 3=6.$$
A: $$2^n\equiv(-1)^n\pmod3 $$
$$2^{n+1}\equiv2(-1)^n\pmod6$$
A: COMMENT.-In the ring $\mathbb Z/6\mathbb Z$ one has $2^3=2$ therefore
$$2^{287}=2^{3\cdot95+2}=2^{95+2}=2^{3\cdot32+1}=2^{3\cdot11}=2^{3\cdot3+2}=2^{3+2}=2\cdot2^2=2^3=2$$
