jmc

Note that $13^{51}$ is $(10+3{)}^{51}$. Any term in the expansion that involves the 10 will be 0 modulo 5, so it suffices to consider $3^{51}(\mathrm{mod}5)$. We look for a pattern in powers of 3. $$\begin{array}{rl}{3}^1& \equiv 3(\mathrm{mod}5)\\ 3^2& \equiv 4(\mathrm{mod}5)\\ 3^3& \equiv 2(\mathrm{mod}5)\\ 3^4& \equiv 1(\mathrm{mod}5).\end{array}$$Since $3^4\equiv 1(\mathrm{mod}5),$ we see that $3^{51}\equiv 3^3\cdot (3^4{)}^{12}\equiv 2(\mathrm{mod}5),$ hence our desired remainder is $\boxed{2}.$