jmc

We are looking at powers of 2, so we notice that $2^3=8=7+1$. Therefore $$2^3\equiv 1(\mathrm{mod}7).$$ In particular $$2^{87}\equiv 2^{3\cdot 29}\equiv 8^{29}\equiv 1^{29}\equiv 1(\mathrm{mod}7).$$ Therefore $$2^{87}+3\equiv 1+3\equiv 4(\mathrm{mod}7).$$ The remainder upon division by 7 is $\boxed{4}$.