Optimizing Diwali Lights problem in C
As @vnp has pointed out, each bulb can have two states, so the number of possibilities is \$2^N\$. However, the state where all bulbs are off is prohibited, so subtract one, and is the answer is simply \$(2^N -1) \mod 10^5\$.

The tricky part of the problem is that \$2^{{10}^4} -1\$ is a very large number to work with, so calculating \$(2^N -1) \mod 10^5\$ is a non-trivial problem.