Find the least positive residue of $10^{515}\pmod 7$. I tried it, but being a big number unable to calculate it.

All answers here use Fermat's little theorem, which is cool, but there is another simpler way (simpler if you don't know Fermat's theorem, otherwise it's more complicated).
$$10^0\equiv 1\mod 7\\
10^1 \equiv 3\mod 7\\
10^2\equiv 10\cdot 10 \equiv 3\cdot 3\equiv 9\equiv 2\mod 7\\
10^3 \equiv 10\cdot 10^2\equiv 3\cdot 2\equiv 6\mod 7\\
10^4\equiv 10\cdot 10^3\equiv 3\cdot 6\equiv 18\equiv 4\mod7\\
10^5\equiv 10\cdot 10^4\equiv 3\cdot 4 \equiv 12\equiv 5\mod 7\\
10^6\equiv 10\cdot 10^5\equiv 3\cdot 5\equiv 15\equiv 1\mod 7$$
Therefore, you know that taking $10^n\mod 7$ will result in a sequence which has repeating values: $1,3,2,6,4,5$, i.e. every sixth (so the zeroth, sixth, twelfth,...) number is $1$. Since $510=6\cdot 85$, you know that the $510$-th number in the sequence will again be $1$, then $511$-th $3$ and so on, so the $515$-th will be $5$.
