smallest positive integer $r$ so that $5^{33333}≡ r (mod 11)$ Title is pretty self explanatory. I tried using the division algorithm to get some a hint about the $5^{33333}$. This was not helpful.
 A: First check that $5^{10} \equiv 1 \pmod{11}$. Now divide the exponent to get $33333 \equiv 3 \pmod{10}$. 
This gives $$5^{33333} \equiv 5^{3} \equiv 4 \pmod{11}.$$
A: $$5\equiv 5\mod 11\\
5^2=25\equiv 3\mod 11\\
5^3= 5^2\cdot 5\equiv 3\cdot 5=15\equiv 4\mod 11\\
5^4= 5^3\cdot 5\equiv 4\cdot 5=20\equiv 9\mod 11\\
5^5= 5^4\cdot 5\equiv 9\cdot 5=45\equiv 1\mod 11\\
$$
So, you know that $5^5\equiv 1\mod 11$.
You also know that $33333=5\cdot 6666+3$
A: Applying standard Congruence Product and Power Rules yields
$$\begin{align}{\rm mod}\,\ 11\!:\,\ \  5^{\large 3+10N} 
&\equiv\, 5\cdot  \color{#0a0}{5^{\large 2}} (\color{#c00}{5^{\large 10}})^{\large N}\\
&\equiv\, 5\cdot\color{#0a0}{3}\  (\color{#c00}1)^{\large N}\ \ \,{\rm by}\ \ \color{#c00}{5^{\large 10}\equiv 1}\ \ {\rm by}\ \, \rm little\ Fermat \\
&\equiv\ \ \ \ \ 4\cdot 1\quad\ \ {\rm and\ \ }\ \color{#0a0}{5^{\large 2}\equiv 3}
\end{align}$$
A: By Fermat's Little Theorem
We have that:
$$a^{11}\pmod{11} \equiv a$$ for any $a$.
So, we have that:
$$5^{11}\equiv 5\pmod{11}$$
Or, more simply we have that:
$$5^{10}\equiv 1\pmod{11}$$
So, we have that:
$$5^{33333}\equiv 5^{3330}\times 5^3\pmod{11}\equiv 5^3\pmod{11} \equiv 4\pmod{11}$$
