What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator 
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?

I did the following:
$$\begin{array}
  & 5 \ \text{mod} \ 7=& &5 \\
  55  \ \text{mod} \ 7= & &6 \\
  555 \ \text{mod} \ 7= & &2 \\
  5555 \ \text{mod} \ 7= & &4 \\
  55555 \ \text{mod} \ 7= & &3 \\
  555555 \ \text{mod} \ 7= & &0 \\
  5555555 \ \text{mod} \ 7= & &5 \\
  55555555 \ \text{mod} \ 7= & &6 \\
  555555555 \ \text{mod} \ 7= & &2 \\
  5555555555 \ \text{mod} \ 7= & &4 \\
\end{array}$$
It can be seen that the cycle is: $\{5,6,2,4,3,0\}$. 
$$\begin{array}
  & 1 \ \text{number =} &5 \\
  7 \ \text{numbers =} &5 \\
  13 \ \text{numbers =} &5 \\
  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots & \\
  985 \ \text{numbers =} &5 \\
  991 \ \text{numbers =} &5 \\
  997 \ \text{numbers =} &5 \\
  998 \ \text{numbers =} &6 \\
  999 \ \text{numbers =} &2 \\
  \color{red}{1000} \ \color{red}{\text{numbers =}} &\color{red}{4} \\
\end{array}$$
From here, we can conclude that $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7 = 4$.
However, I wasn't allowed to use a calculator and solved this in about 12 minutes. Another problem was that there was a time limit of about 5 minutes. My question is: Is there an easier and faster way to solve this?
Thanks a lot in advance!

After noting $555555$ is divisible by $7$, note further that $555555\times 10^r$ is divisible by $7$ for any positive integer $r$. So you can cast out groups of six $5$s starting at the most significant digit, without changing the remainder on division by $7$. This gets rid of $996$ of the $5$s, leaving $5555$. Then $4949$ is obviously divisible by $7$ leaving $606$, and simple division then gives the remainder $4$.
