imc

Notice that $L_{17}$ contains the highest power of every prime below $17$ since higher primes cannot divide $L_{17}$. Thus, $L_{17}=16\cdot 9\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17$. When writing the sum under a common fraction, we multiply the denominators by $L_{17}$ divided by each denominator. However, since $L_{17}$ is a multiple of $17$, all terms will be a multiple of $17$ until we divide out $17$, and the only term that will do this is $\frac{1}{17}$. Thus, the remainder of all other terms when divided by $17$ will be $0$, so the problem is essentially asking us what the remainder of $\frac{L_{17}}{17}=L_{16}$ divided by $17$ is. This is equivalent to finding the remainder of $16\cdot 9\cdot 5\cdot 7\cdot 11\cdot 13$ divided by $17$. We use modular arithmetic to simplify our answer: This is congruent to $-1\cdot 9\cdot 5\cdot 7\cdot 11\cdot 13(\mathrm{mod}17)$. Evaluating, we get: $$\begin{array}{rl}(-1)\cdot 9\cdot 35\cdot 11\cdot 13& \equiv (-1)\cdot 9\cdot 1\cdot 11\cdot 13(\mathrm{mod}17)\\ & \equiv 9\cdot 11\cdot (-13)(\mathrm{mod}17)\\ & \equiv 9\cdot 11\cdot 4(\mathrm{mod}17)\\ & \equiv 2\cdot 11(\mathrm{mod}17)\\ & \equiv 5(\mathrm{mod}17)\end{array}$$ Therefore the remainder is $\boxed{5}$.