<s>[INST] 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!
 [/INST] There are some great answers here, but If your doing mental math, this might be the easiest if you basic modular arithmetic properties. 
First look at the pattern 5, 55, 555, 555.... If you notice that this can be written as 
$a_1 = 5$ and $a_n = 10a_{n-1}+5$ 
This produces the pattern. Now if we perform the modulo we get 
$a_n \equiv 3a_{n-1}-2~\mod 7$. 
This is a lot easier to work with. We can then do the same method you did, calculating the cycle:
$5 \equiv a_1 \equiv 5 \mod 7$
$55 \equiv 3(5)-2 \equiv 6 \mod 7$
$555 \equiv 3(6)-2 \equiv 2 \mod 7$
$5555 \equiv 3(2)-2 \equiv 4 \mod 7$
$55555 \equiv 3(4)-2 \equiv 3 \mod 7$
$555555 \equiv 3(3)-2 \equiv 0 \mod 7$
$5555555 \equiv 3(0)-2 \equiv 5 \mod 7$
Therefore the cycle is, as you said ${(5,6,2,4,3,0)}$ Now
 $1000 \equiv 10(10)(10) \equiv 4(4)(4) \equiv 4(16) \equiv 4(4) \equiv 4 \mod 6$.
 So we pick the $4$th element, therefore the answer is $4$
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