jmc

We can look at the terms of the Lucas sequence modulo 8. $$\begin{array}{rl}{L}_1& \equiv 1(\mathrm{mod}8),\\ L_2& \equiv 3(\mathrm{mod}8),\\ L_3& \equiv 4(\mathrm{mod}8),\\ L_4& \equiv 7(\mathrm{mod}8),\\ L_5& \equiv 3(\mathrm{mod}8),\\ L_6& \equiv 2(\mathrm{mod}8),\\ L_7& \equiv 5(\mathrm{mod}8),\\ L_8& \equiv 7(\mathrm{mod}8),\\ L_9& \equiv 4(\mathrm{mod}8),\\ L_{10}& \equiv 3(\mathrm{mod}8),\\ L_{11}& \equiv 7(\mathrm{mod}8),\\ L_{12}& \equiv 2(\mathrm{mod}8),\\ L_{13}& \equiv 1(\mathrm{mod}8),\\ L_{14}& \equiv 3(\mathrm{mod}8),\\ L_{15}& \equiv 4(\mathrm{mod}8),\\ L_{16}& \equiv 7(\mathrm{mod}8).\end{array}$$Since $L_{13}=1$ and $L_{14}=3$, the sequence begins repeating at the 13th term, so it repeats every 12 terms. Since the remainder is 4 when we divide 100 by 12, we know $L_{100}\equiv L_4(\mathrm{mod}8)$. Therefore the remainder when $L_{100}$ is divided by 8 is $\boxed{7}$.