Sum of powers of $2$ modulo $11$: $\sum_{i=0}^{5 \cdot x -1} [2^{4i}] \equiv 0 \mod 11$ I tried the proof by reducing the exponents modulo 10, but didn't really get a solution :/.
Would love some help :D, thanks guys
 A: Observe that
$$\sum_{i=0}^{5x-1}2^{4i}=\sum_{i=0}^{5x-1}16^i=\sum_{j=0}^{x-1}\sum_{k=0}^416^{5j+k}=\sum_{j=0}^{x-1}\sum_{k=0}^4(16^{5j}\cdot16^k)=\sum_{j=0}^{x-1}16^{5j}\sum_{k=0}^416^k.$$
Now working modulo 11, note that $16\equiv 5\bmod 11$, so that
$$\sum_{k=0}^416^k\equiv\sum_{k=0}^45^k=1+5+25+125+625\equiv 1+5+3+4+9=22\equiv 0\bmod 11$$
and thus
$$\sum_{i=0}^{5x-1}2^{4i}=\sum_{j=0}^{x-1}16^{5j}\sum_{k=0}^416^k\equiv \sum_{j=0}^{x-1}(16^{5j}\cdot 0)\equiv 0\bmod 11.$$
In other words,
$$\begin{array}{c|c|c|c|c|c|c|c|c}
i\strut & 0 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots\\\hline
2^{4i}\bmod 11\strut & 1 & 5 & 3& 4 & 9 & 1 & 5 & \cdots\\\hline
\end{array}\\\qquad\qquad\qquad\quad\;\;\underbrace{\hspace{1.2in}}_{\large\text{sum }\equiv\, 0\,\bmod 11}\underbrace{\hspace{1.2in}}_{\large \text{repeats every 5}}$$
A: Use the fact that in genral
$$(1-a)(1+a+a^2+\cdots +a^{n-1})=1-a^n.\tag{1}$$
Let $a=2^4$ and $n=5x$. 
By (1), if $S$ is our sum, then $(1-a)S=1-2^{20x}$.  But by Fermat's Theorem, we have $2^{10}\equiv 1\pmod{11}$, and hence $2^{20x}\equiv 1\pmod{11}$ for all $x$.
It follows that $S\equiv 0\pmod{11}$ for all $x$.  
A: ${\rm mod}\ 11\!:\ a=2^4\,\Rightarrow \color{#0a0}{a^5 =} (2^{10})^2\overset{\rm Fermat}\equiv\! 1^2\color{#0a0}{\equiv 1}.\,$ Thus, with $\,x = k\!+\!1,\,$ so $\,5x\!-\!1 = \color{#c00}{5k\!+\!4},\,$
summing a geometric series: $ $ mod $\smash[b]{\,11\!:\ \underbrace{1+a+\cdots+a^4}\, =\, \dfrac{\color{#0a0}{a^5-1}}{a-1}\equiv \color{#c0f}0\ \ {\rm by}\ \ \color{#0a0}{a^5\equiv 1},\,\ a\not\equiv 1}$ 
$\quad\ \ \begin{eqnarray}\Rightarrow\ \ 1+a+\cdots + a^{\color{#c00}{5k+4}}\!\!\!\!\! &=&\quad\,\ \overbrace{(1+a+\cdots +a^4)}^{\large \color{#c0f}{\equiv\ 0}}\\ &\qquad\ +&\,\ a^5(1+a+\cdots+a^4)\\ &\qquad\,\ +\,&a^{10}(1+a+\cdots+a^4)\\ &\qquad\ \,&\qquad\qquad\ \ \vdots\\  &\qquad\ +& a^{\color{#c00}{5k}}(1+a+\cdots+a^{\color{#c00}{4}} )\\ &\equiv&\quad\ \ \color{#c0f}0\end{eqnarray}$
