What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator

${\rm mod}\ 7\!:\,\ \overbrace{55\cdots 55}^{1+3n\rm\,\ fives}\, =\, \dfrac{5(10^{1+3n}\!-1)}9\, \equiv\, \dfrac{-2\,(3^{1+3n}-1)}2 \,\equiv\, -3(\color{#c00}{3^3})^{n}\!+1 \equiv 4\ $ by $\ \color{#c00}{3^3\equiv -1},\ n$ odd

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 $4$.

Note that $111111$ is divisible by $7$. This follows either from $111111 = \frac{10^6-1}{9}$ where $10^6-1$ is divisible by $7$ by Fermat's little theorem, or from the factorization $111111 = 111 \cdot 1001 = 111 \cdot 7 \cdot 11 \cdot 13$. This implies that $555555$ is also divisible by $7$, hence $$ \underbrace{555 \cdots 5}_{k \text{ times } 5} \mod 7 $$ is periodic with period $6$: if $k = 6a+b$, then $$ \underbrace{555 \cdots 5}_k = \underbrace{5555}_b + 555555 \cdot 10^b + 555555 \cdot 10^{6+b} + \cdots + 555555 \cdot 10^{6(a-1)+b} $$ where all terms (expect the first) are divisible by $7$. You can now draw the desired conclusion, after noting that $1000 \equiv 4 \mod 6$.