jmc

Replacing each term in the sum by a modulo 15 equivalent, we have $$\begin{array}{rl}& 1+6+11+16+21+26+\cdots +91+96+101\\ & \equiv 1+6+11+1+6+11+\cdots +1+6+11(\mathrm{mod}15),\end{array}$$where the terms $1+6+11$ are repeated $7$ times on the right.

Since $1+6+11=18\equiv 3(\mathrm{mod}15)$, we have $$\begin{array}{rl}1+6+11& +16+21+26+\cdots +91+96+101\\ & \equiv \underset{3}{\underbrace{1+6+11}}+\underset{3}{\underbrace{1+6+11}}+\cdots +\underset{3}{\underbrace{1+6+11}}\\ & \equiv 7\cdot 3\\ & =21\\ & \equiv \boxed{6}(\mathrm{mod}15).\end{array}$$