jmc

To find the remainder when $5^{999999}$ is divided by 7, we look at the first few powers of 5 modulo 7: $$\begin{array}{rl}{5}^0& \equiv 1,\\ 5^1& \equiv 5,\\ 5^2& \equiv 5\cdot 5\equiv 25\equiv 4,\\ 5^3& \equiv 5\cdot 4\equiv 20\equiv 6,\\ 5^4& \equiv 5\cdot 6\equiv 30\equiv 2,\\ 5^5& \equiv 5\cdot 2\equiv 10\equiv 3,\\ 5^6& \equiv 5\cdot 3\equiv 15\equiv 1(\mathrm{mod}7).\end{array}$$Since $5^6\equiv 1(\mathrm{mod}7)$, the remainders become periodic, with period 6. Since $999999\equiv 3(\mathrm{mod}6)$, $5^{999999}\equiv 5^3\equiv \boxed{6}(\mathrm{mod}7)$.