smallest positive integer $r$ so that $5^{33333}≡ r (mod 11)$

Question

Title is pretty self explanatory. I tried using the division algorithm to get some a hint about the $5^{33333}$. This was not helpful.

Answer 1

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: $$5^{33333}\equiv 5^{3330}\times 5^3\pmod{11}\equiv 5^3\pmod{11} \equiv 4\pmod{11}$$

Answer 2

First check that $5^{10} \equiv 1 \pmod{11}$. Now divide the exponent to get $33333 \equiv 3 \pmod{10}$. This gives $$5^{33333} \equiv 5^{3} \equiv 4 \pmod{11}.$$

Answer 3

$$5\equiv 5\mod 11\\ 5^2=25\equiv 3\mod 11\\ 5^3= 5^2\cdot 5\equiv 3\cdot 5=15\equiv 4\mod 11\\ 5^4= 5^3\cdot 5\equiv 4\cdot 5=20\equiv 9\mod 11\\ 5^5= 5^4\cdot 5\equiv 9\cdot 5=45\equiv 1\mod 11\\ $$

So, you know that $5^5\equiv 1\mod 11$.

You also know that $33333=5\cdot 6666+3$

Answer 4

Applying standard Congruence Product and Power Rules yields

$$\begin{align}{\rm mod}\,\ 11\!:\,\ \ 5^{\large 3+10N} &\equiv\, 5\cdot \color{#0a0}{5^{\large 2}} (\color{#c00}{5^{\large 10}})^{\large N}\\ &\equiv\, 5\cdot\color{#0a0}{3}\ (\color{#c00}1)^{\large N}\ \ \,{\rm by}\ \ \color{#c00}{5^{\large 10}\equiv 1}\ \ {\rm by}\ \, \rm little\ Fermat \\ &\equiv\ \ \ \ \ 4\cdot 1\quad\ \ {\rm and\ \ }\ \color{#0a0}{5^{\large 2}\equiv 3} \end{align}$$