$6^{66}\equiv r \pmod {66}$ The answer doesn't need to be exact, the possible answers to the exercise are "between 30 and 40", "from 50 to 66" or something akin to that.
I've no idea how to solve this. Previous problems of this kind were simple usage of Euler's theorem but I don't see how I can apply it in this case.
 A: From Euler's theorem or Fermat's little theorem, we have
$$6^{10} \equiv 1 \pmod{11} \implies 6^{66} \equiv 6^6 \pmod{11} \implies 6^{66} \equiv 5 \pmod{11}$$
This means
$$6^{66} \equiv 5,16,27,38,49,60 \pmod{66} \,\,\,\, (\spadesuit)$$
Further, we have
$$6^{66} \equiv 0 \pmod6 \,\,\,\, (\heartsuit)$$
From $(\spadesuit)$ and $(\heartsuit)$, we have
$$6^{66} \equiv 60\pmod{66}$$
A: You can do this a bit by brute force trying to hit small numbers. This is bound to work reasonably well because there are only $11$ possible values for the answer.
$$6^3=216\equiv 18\bmod 66$$ $$6^6\equiv18^2=324\equiv -6 \bmod 66$$ $$6^{66}\equiv (-6)^{11}\equiv -6^{11}\equiv 6^6\equiv -6 \equiv 60$$
A: Looking at early powers of $6 \bmod 66$  you can see that $$6^{11} \equiv 6\bmod 66$$ and $$6^6 \equiv 60\bmod 66$$
So you can say $$6^{66} = \left(6^{11}\right)^6 \equiv 60 \bmod 66$$ 
A: Notice $\,\ {\rm mod}\ 11\!:\ \ \color{#c00}{6^5\,\equiv\, -1}\,\ $ by $\,\ 6(6^2)^2\equiv 6(3)^2\equiv 6(-2)\equiv\color{}{ -1}$ 
Therefore $\ 6^{65}\equiv (\color{#c00}{6^5})^{13}\equiv (\color{#c00}{-1})^{13}\equiv\color{#0a0}{ -1},\,$ so $\,6^{65} = 11n\color{#0a0}{-1}\,\overset{\times\,6}\Rightarrow\, 6^{66} = 66n-6$
