jmc

We can look at the terms of the Fibonacci sequence modulo 8. $$\begin{array}{rl}{F}_1& \equiv 1(\mathrm{mod}8),\\ F_2& \equiv 1(\mathrm{mod}8),\\ F_3& \equiv 2(\mathrm{mod}8),\\ F_4& \equiv 3(\mathrm{mod}8),\\ F_5& \equiv 5(\mathrm{mod}8),\\ F_6& \equiv 0(\mathrm{mod}8),\\ F_7& \equiv 5(\mathrm{mod}8),\\ F_8& \equiv 5(\mathrm{mod}8),\\ F_9& \equiv 2(\mathrm{mod}8),\\ F_{10}& \equiv 7(\mathrm{mod}8),\\ F_{11}& \equiv 1(\mathrm{mod}8),\\ F_{12}& \equiv 0(\mathrm{mod}8),\\ F_{13}& \equiv 1(\mathrm{mod}8),\\ F_{14}& \equiv 1(\mathrm{mod}8),\\ F_{15}& \equiv 2(\mathrm{mod}8),\\ F_{16}& \equiv 3(\mathrm{mod}8).\end{array}$$Since $F_{13}$ and $F_{14}$ are both 1, 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 $F_{100}\equiv F_4(\mathrm{mod}8)$. Therefore the remainder when $F_{100}$ is divided by 8 is $\boxed{3}$.